Unit 2  Quantities in Chemistry

#### Atomic Mass

The sum of individual atoms can be used to find the mass of a molecule.

The mass of hydrogen peroxide, H2O2 would be calculated like this:

H2O2 has 2 hydrogen atoms and 2 oxygen atoms in it.

Therefore the mass is
2 X H   =  2 X 1.01 u   = 2.02 u

2 X O  = 2 X 16.00 u = 32.00 u
34.02u

So one molecule of hydrogen peroxide weighs in at 34.02 u.

The periodic table provides you with individual atomic masses.  If you know the number and type of elements in a molecule you can add up the individual masses to find the molecular mass or molecular weight.

Find the molecular mass of calcium phosphate, Ca3(PO4)2

The molecule has 3 calcium atoms, 2 phosphate atoms and 8 O atoms in it. Stop and verify this for yourself. The Ca has a subscript 3 with it. The P has an assumed 1 and the O has a 4. However the PO4  group has a set of brackets around it with a subscript 2. The 2 means multiply everything inside the brackets by 2. So we end up with the 2 P and 8 O atoms.

Calculation:  3 X Ca = 3 X 40.08 u = 120.24 u

2 X P   = 2 X 30.97 u =   61.94 u

8 X O  = 8 X 16.00 u = 128.00 u

310.18 u

One molecule of calcium phosphate weighs 310.18 u

Stop here and do the  Atomic Masses Exercise

Percent Composition

Percent composition is a simple little calculation that has a large impact on chemistry. When a brand new substance is discovered one of the first things that is determined is its chemical make-up. That means how much of each type of atom is in the molecule. What is the percent composition of strontium oxide? SrO

There is 1 atom of Sr and 1 atom of O.

The mass of 1 Sr is 87.62 u and the mass of an O is 16.00 u.

The percent composition is the fraction of the SrO that is just Sr.  This fraction is:

Percentage of Sr = mass of Sr in the molecule / molecular mass X 100%

= 87.62 u / 103.62 u X 100%

= 0.8456 X 100%

=  84.56%

Strontium makes up 84.56% of the molecule. The oxygen makes up 100% - 84.56% = 15.44%. The u's are units that cancel out. The % sign means out of 100.

The general equation for finding the percentage composition of an element is:

percentage of an element = the total mass of just that element
molecular mass X 100%

#### Stop here and do the  Percent Composition Exercise

Homework  Section 2.2  Questions  Page 106    Answers

 Molecular Amounts versus Labratory Amounts The Mole: The Start of Chemical Calculations The creation of formulas and the finding of their molecular masses is only partially complete. What exactly does that number you have found for the molecular mass mean?  Is it the mass of one molecule, or a million, or ten billlion?  What unit is it measured in? The Carbon-12 Based Atomic Mass Scale In the years following Dalton's presentation of his atomic theory, chemists worked very hard at determining a complete set of relative masses for all the elements that were known.  By knowing these relative masses the chemists were able to select amounts of elements in grams needed for any desired atom ratio.    In establishing a table of atomic amsses, it is necessary to have a reference point against which to compare the relative masses. Currently, the agreed-upon reference is the most abundant isotope of carbon, which is carbon-12. By definition, an atom of this isotope is defined as having the mass of exactly 12.000 u. (atomic mass units)  In other words, an amu is defined as 1/12th of the mass of one atom of carbon-12. The definition of the size of the atomic mass unit was quite arbitrary.  It could just as easily have been selected to be 1/24th of the mass of one atom of a carbon atom, or 1/10th the mass of a calcium atom, or any other value. Why 1/12th the mass of carbon-12?  Carbon is a very common element, available to any scientist  and by choosing the amu to be of this size, the atomic masses of nearly all the other elements are almost whole numbers, with the lightest atom having a mass of approximately 1. (hydrogen-1 has a mass of 1.007825 amu when carbon-12 is assigned a mass of exactly 12 amu.) The Mole Suppose we want to make molecules of carbon dioxide, CO2, in such as way that there would be no extra carbon and oxygen atoms left. If we took ten atoms of carbon and twenty atoms of oxygen we would make 10 molecules of carbon dioxide.  Suppose we wanted to make more, lets say 40,000 molecules of carbon dioxide.  Once again, we could count out 40,000 atoms of carbon and 80,000 atoms of oxygen, let them react and we'd have 40,000 molecules of carbon dioxide.  This sounds all very nice and neat unless you realize the trap.  Have you ever seen an atom?  Atoms are too tiny to count individually.  Saying they are tiny is even wrong because tiny can be seen.   Even with the best scanning tunneling electron microscope ever invented the largest atoms known, look just like fuzzy cloud tops.  We have never seen individual atoms.  There are no lenses with the resolving power or balances fine enough to measure an individual atom. We get around this problem because each element has its own characteristic atomic mass and each formula has its own unique molecular mass.    We know, that oxygen weighs in at 1.33 times that of carbon, because of this we get a ratio of their masses:                    16.0 amu (for one atom of O) = 1.33                    12.0 amu (for one atom of C)       1 If we take a sample of oxygen and carbon in a ratio of 1.33 to 1, we must obtain equal numbers of their atoms.  That is, if we actually had a balance that could measure amu directly we could mass out  32 amu of oxygen and 12 amu's of carbon - a mass ratio of 2.66 to 1 - we would have exactly 2 atoms of oxygen for every atom of carbon and a 2 to 1 ratio by atoms. When we mass out a sample of an element such that its mass in grams is numerically equal to the element's atomic weight, we always obtain the same number of atoms no matter what element we choose.  Thus 12.0 g of carbon has the same number of atoms as 16.0 grams of oxygen, or 32.1 g of sulphur, or 55.8 g or iron. This relationship also extends to compounds. The formula mass of water, H2O, is 18.0 amu. If we take 18.0 grams of water then it should have the same number of molecules as there were atoms in 12.0 grams of carbon.  The carbon-12 isotope, which makes up 98.89% of all naturally occuring carbon, is the reference used by SI Metric in its definition of the base unit for a chemical substance, the mole, abbreviated mol. This mole concept is the most important in all of chemistry. Once this concept is grasped all the rest of chemistry will appear easy. Avogadro's Number The mole is defined as 6.02 x 1023 units.  It is called Avogadro's number in honour of Italian scientist, Amadeo Avogadro (1776-1856). It is a pure number with a special name, just like so many others.  For example:                                                  2 = pair                                                12 = dozen                                              144 = gross                                              500 = ream                                 6.02 X 1023 = Avogadro's number This number is not an odd number at all. It became inevitable once the amu was defined.   The relationships needed are:                                           1 mole = 6.02 X 1023 particles     (General expression)                                                          or                                          12 amu = 1 atom of  C      (Specific example)                                     1 mol of C = 6.02 X 1023 atoms of C                                     1 mol of C = 12 g of  C Another useful relationship is that 1 amu = 1.66 X 10-24 g SO                            1  amu               =  6.0 X 1023 particles                       1.66 X 10-24  grams The mass in grams of a substance, that equals one mole is often called its molar mass, and the units are grams/mole or g/mol.   For example, aspirin has a molecular mass of 180 grams. Therefore if we massed out exactly 180 grams of aspirin we would have Avogadro's number of aspirin molecules. Stop here and do the Atoms and Moles Exercise Sheet

 Using the Mole Concept One mole of any substance can be calculated from its formula mass. Since this is true it is absolutely essential that when you are using the mole concept that the correct formula be used. It is not enough to say "use 1 mole of nitrogen".  Do we mean atomic elemental nitrogen or nitrogen gas?  There is a difference!  One mole of  N consists of Avogadro's number of nitrogen atoms (and has a mass of 14.01 g), whereas 1 mole of N2 consists of Avogadro's number of molecules, each molecule having two nitrogen atoms. One mole of N2 molecules would have a mass of 2 X 14.01 g = 28.02 g. One of the advantages of the mole concept is that it lets us think about formulas on two levels at the same time.  One level is that of atoms or molecules or ions, and the other level is that of lab-sized practical quantitites, such as moles and grams.  Look at the equation below: H2O   consists of                   2 H                               +    O 1 molecule of H2O                    2 atoms of H                      1 atom of O 1 dozen H2O molecules            2 dozen atoms of H            1 dozen O atoms 6.02 X 1023 H2O molecules     12.04 X 1023 H atoms        6.02 X 1023 O atoms 1 mole of H2O molecules         2 moles of H atoms            1 moles of O atoms 18.0 g of H2O                            2.0 g of H                           16.0 g O atoms When we think about H2O at the first level, we can easily see that a dozen of its molecules are made from two dozen atoms of H and one dozen atoms of O.  However, if we switch to the more practical lab-sized level, it is just as easy to think about one mole of H2O and to view this quantity as consisting of two moles of H and one mole of O. The numbers in all but the last row are in the same ratio regardless of the scale, whether we deal with single particles or with moles of them. After planning an experiment at the mole level, it is easy to convert numbers of moles into corresponding masses of chemicals to meet any desired needs. Extension Exercise 2.1 Ratios in Chemical Equations Extension Exercise 2.1 Amounts in Chemistry

 Converting Moles to Grams This is probably the single most used equation in chemistry.  It is the one that allows the conversion of moles into grams and grams back into moles.  Moles are a "theoretical value" which looks good on paper.  Grams is the "practical value" that you take out of a stock bottle and place on the balance.  Lab balances read in grams, not moles, and so in practical work all mole values have to first be converted. Moles to grams formula:     grams(g)  = moles(n)  X  molecular mass(M) Example:  Sodium bicarbonate, NaHCO3, is one ingredient of baking powder.  How many grams of sodium bicarbonate are in 0.673 moles? Step 1.  Find the molecular mass:           NaHCO3 = 1 Na = 1 X 22.99 g/mol =  22.99 g/mol                              1 H   = 1 X   1.01 g/mol =    1.01 g/mol                                1 C = 1 X 12.01 g/mol =  12.01 g/mol                                3 O = 3 X 16.00 g/mol =  48.00 g/mol                                                                         84.01 g/mol Step 2. Use the equation:   g = n  X  M                                             = 0.673 mol X 84.01   g                                                                               mol                                             = 56.54 g Step 3: Provide a written answer:  There are 56.54 grams of sodium bicarbonate in 0.673 moles. Grams to moles formula:    n =     g                  M Example:  Potassium permanganate, KMnO4, at one time was used for an anti-fungal agent. You                 could always tell someone who had just been treated because their feet were purple. The                 pharmacy gives you 250 grams of the stuff in a bottle.  How many moles of it do you                 have? Step 1:  Find the molecular mass of KMnO4:               1  K  = 1 X 39.10 g/mol  =  39.10 g/mol               1 Mn = 1 X 54.94 g/mol  =  54.94 g/mol               4 O   =  4 X 16.00 g/mol  = 64.00 g/mol                                                        158.04 g/mol Step 2:  Use the equation:  n =           g                    molecular mass =       250 g             158.04 g/mol =  1.58 mol Step 3:  Write a written answer. A sample of 250 grams is equivalent to 1.58 moles of potassium permanganate. Finding the molecular mass In the above two examples you have had a molecular formula to work with.  As long as you have a molecular formula, or name, from which you can make a formula up, you will always have a molecular mass.   In order to find the molecular mass, without a name to go on, you need two pieces of information. The number of grams and the number of moles that it represents. molecular mass(M) =    grams     =        g                                          moles               n Example:   A fellow student comes to you with a sample of an unknown chemical. They tell you that it has a mass of 34.91 grams and that it is exactly 0.20 moles.   What is the molecular mass of the substance? Step 1:  Use the equation:     molecular mass =   34.91  g   =  174.55     g                                                                                 0.20 mol                    mol Step 2: Write a sentence.  The unknown substance has a molecular mass of 174.55 g/mol. Stop here and go to the Grams, Moles and Molar Mass Worksheet Converting Moles into Molecules The formula needed is:     molecules = n * NA Multiply the number of moles by Avogadro's number to find the number of molecules. This is similar to multiplying 4 cartons of eggs by 12 eggs/carton to find the total number of eggs present. Ex.   How many molecules are there in 3 moles of CO2?                 molecules = n * NA                            = 3 moles * 6.0 X 1023 molecules/mole                            = 18.06 X 1023 molecules                            = 1.806 X 1024 molecules There are 1.806 X 1024 molecules in 3 moles of CO2. Converting Molecules into Moles The necessary formula is:     moles (n)  = molecules                                                                           NA This concept is similar to counting 144 eggs and dividing by 12 to get the number of cartons. Example: You have a sample of 5.6 X 1024 molecules of XeF6.  How many moles of this particle sample of XeF6 do you have? moles =    molecules                       NA                =        5.6 X 1024 molecules                       6.02 X 1023 molecules/mole            =   9.3 moles There are 9.3 moles of XeF6 in this sample. Converting Molecules into Atoms You have to be able to look inside a molecule and determine the total number of atoms in its makeup. This is the atoms subscript total.  For example H2O has a subscript total of 3 atoms.  H2SO4 has a subscript total of 7 atoms.  MgSO4•7 H2O has a subscript total of 27 atoms. The formula needed is:    atoms = molecules X subscript total Example: You have 1.2 X 1023 molecules of hydrogen peroxide, H2O2.  How many atoms in total do you have? atoms = molecules X subscript total            = 1.2 X 1023 molecules X 4 atoms/molecule            = 4.8 X 1023 atoms There are 4.8 X 1023 atoms in total in this sample of H2O2. Converting Atoms into Molecules The formula needed is:   molecules =    total atoms                                                                  subscript total Exmaple:  You have a sample of calcium hydroxide, Ca(OH)2.   You have a total of 5 million atoms. How many molceuls are presnet? molecules =    total atoms                        subscript total                  =    5,000,000 atoms                         5 atoms/molecule                  =   1,000,000 molecules There are 1 million molecules in this sample of Ca(OH)2. Extension Exercise 2.2 Calculations Involving the Mole Stop here and do the Marvin Da Mole Worksheet Stop here and do the Marvin Da Mole Strikes Again Worksheet Determining Chemical Formulas (Empirical Formulas)

 When the quantitative analysis of a compound gives us the masses of each element in the same-size sample, we can convert their masses into proportions by moles by a grams to moles calculation.  With atomic particles, proportions by moles are numerically the same as proportions by atoms, which are just the numbers we need to construct a chemical formula.  The formula obtained in this way is called an empirical formula.  An empirical formula is the one that uses as subscripts the smallest whole numbers that describe the ratios of atoms in a compound. Calculating an Empirical Formula from data obtained by quantitative analysis. Example #1 A sample of an unknown compound with a mass of 2.571 grams was found to contain 1.102 grams of C and 1.469 grams of oxygen.  What is its empirical formula? 1.102 g of C X 1 mole of C  = 0.09176 moles of C                         12.01 g of C 1.469 g of O X 1 mole of O = 0.09181 moles of O                          16.00 g of O Therefore the formula is C0.09176O0.09181   Pick the smallest of the subscript numbers as the divisor and divide it into all the other numbers.                         C1.0000O1.0001 = C1O1 As a general rule, if the calculated subscript differs from a whole number by only several units in the last place, we can safely round to the whole number. Example #2 When a sample with a mass of 2.448 grams of compound present in liquified petroleum gas was analyzed, it was found to contain 2.003 grams of carbon and 0.4448 grams of hydrogen.   What is its empirical formula? 2.003 g of C X 1 mole of C  = 0.16678 moles of C                         12.01 g of C 0.4448 g of H X 1 mole of H = 0.4404 moles of H                            1.1 g of H Therefore the formula is C0.16678H0.4404 = C1.000H2.667 The ratios don't come out to the expected nice ratios.  The solution to this is to multiply to see if we can get a better whole number ratio. C0.16678 X 2H0.4404 X 2 = C2H5.334             Still not good enough       C0.16678 X 3H0.4404 X 3 = C3H8.001   =  C3H8 Example #3 A 5.438 gram sample of ludlamite, a greyish-blue mineral sometimes used in ceramics, was found to contain 2.549 grams of iron, 1.947 grams of oxygen, and 0.9424 grams of phosphorus.  What is its empirical formula? 2.549 grams of Fe X 1 mole of Fe = 0.0456 moles of Fe                                    55.85 g Fe 1.947 grams of O X 1 mole of O = 0.1217 moles of O                                    16.00 g O 0.9424 grams of P X 1 mole of P = 0.0304 moles of P                                     30.97 g P Therefore the formula is Fe0.0456O0.1217P0.0304 = Fe1.5O4.0P1 =Fe3O8P2 Calculating Empirical Formulas from Percentage Compositions Example #1 Barium carbonate, a white powder used in paints, enamels, and ceramics, has the following composition: Ba, 69.58%; C, 6.090%; O, 24.32%.  What is its empirical formula? Assume we have a 100 gram sample since it is out of 100 percent.  Therefore all the percentages change to grams and solve it like the above three examples. 69.58  grams of Ba X 1 mole of Ba = 0.5067 moles of Ba                                     137.33 g Ba 6.090 grams of C X 1 mole of C = 0.5071 moles of C                                    12.01 g C 24.33 grams of O X 1 mole of O = 1.520 moles of O                                    16.00 g O Therefore the formula is Ba0.5067C0.5071O1.520 = Ba1C1.001O2.999 = BaCO3 Example #2 Calomel is the common name of a white powder once used in the treatment of syphilis.  Its composition is 84.98% mercury and 15.02% chlorine.  What is its empirical formula? 84.98 grams of Hg X 1 mole of Hg = 0.4237 moles of Hg                                     200.59 g Hg 15.02 grams of Cl X 1 mole of Cl =  0.4237 moles of Cl                                    35.45 g Cl Therefore the formula is Hg0.4237Cl0.4237 = HgCl Combustion Analysis In combustion analysis a sample of the chemical being analysed is burned with oxygen.  The resulting products are used with a balanced combustion equation to stoichiometrically determine the moles of the starting elements.  The mass spectrometer will give you the peercentages of the atoms as they are burned.  You can then use these in a percentage calculation to get your empirical formula. Read Page 108-110  for a more complete description of the mass spectrometer and combustion analayzer. Determining a Molecular Formula from an Empiricial Formula and a Formula Weight Example #1 Styrene, the raw material for polystyrene plastics, has the empirical formula CH. Its formula wieght is 104.08 g/mole.  What is its molecular formula? Find the molecular mass of the empirical formula: the empirical moelcular mass CH = 13.1 g/mole Divide the actual molecular mass by the empirical molecular mass:  104.08 g/mole / 13.1 g/mole = 8 Multiply the empirical formula by this number:  C1H1 X 8 = C8H8 Example #2 Calomel has a formula weight of 472.08.  The empirical formula is HgCl.  What is its actual molecular formula? HgCl = 200.59 + 35.45 = 236.04 g/mole 472.08 g/mole / 236.04 g/mole = 2 Therefore Hg1Cl1 X 2 = Hg2Cl2 Stop here and go to the Empirical Calculations Worksheet Homework Section 2.3 Questions Page 120 Extension Exercise 2.3 Determining Chemical Formulas Investigation 2.4  Percentage Composition by Mass of Magnesium Oxide Alternative Exercise 2.4  Percentage Composition of a Copper Compound Self Quiz 2.1 - 2.4

 Introduction to Solutions and Concentration Whenever possible, reactions are carried out with all of the reactants in the same fluid phase. This means that it is preferrable to have liquids reacting with liquids, or gases reacting with gases. This is because no matter what type of particle we talk about, in order for them to react they must collide with each other.  Solids have little or no movement and therefore offer few opportunities for collision.  Liquids and gases are both fluids and offer much more opportunity for frequent collision.  Liquids are used in the lab because they are fairly easy to create and use.  Gases are used but specialized gas handing equipment is required. A solution is a uniform mixture of particles of atomic, ionic, or molecular size.  A minimum of two substances are present.  One is called the solvent and all the others are called the solutes.  The solvnet is the fluid medium in which all of the solutes are dissolved.  A solvent can be a solid, liquid or a gas but the most common solvent is water and therefore we deal almost exclusively with aqueous solutions.   Solutes are any substance dissolved in the solvent. A dilute solution is one in which the ratio of solute to solvent is very small, for example, a few crystals of sugar in a glass of water.  In a concentrated solution the ratio of solute to solvent is large.  Maple syrup is a concnetrated solution of sugar in water.  A saturated solution is one in which no more solute can be dissolved at a particular temperature.  An unsaturated solution is one in which the ratio of solute to solvent is lower than that of the corresponding saturated solution.  If more solute is added to an unsaturated solution, at least some of it should dissolve.   A supersaturated solution is an unstable system in which the ratio of dissolved solute to solvent is higher than that of a saturated solution.  A supersaturated solution can be made by gently cooling a hot saturated solution.  At a lower temperature the dissolved solute can be made to precipitate out when a seed crystal is added.  The process is called precipitation and the substance that forms is the precipitate. The amount of solute needed to make a saturated solution in a given quantity of solvent at a specific tempeature is called the solubility of the solution.

 Units of Concentration (v/v, w/v and ppm) These units of concentration are most often seen and used with commerical products.  Expect for 'ppm' they are not used often in the lab. Percent Concentration Volume/Volume (v/v):  used with 2 liquids. % Concentration = Vsolute  X 100%                                        Vsolvent     eg.   5 mL of vinegar are dissolved in 100 mL of viniegar solution.  What is its v/v concentration.         % concentration =   5 mL    X 100%  = 5%                                         100 mL Example #1   A photographic stop bath contains 140 mL of pure acetic acid in a 500 mL bottle of solution.  What is the v/v concentration? % concentration = 140 mL X 100% = 38.9%                                         360 mL Weight/Volume (w/v): used with one solid and one liquid This means there is a certain mass, in grams, in every 100 mL of solution.   eg.   a 3% H2O2 topical antibiotic solution means that there is 3 grams of H2O2 in every 100 mL of solution. Parts per Million Concentration (ppm) Environmental solution are often very low in concentration.  We often use terms like: 1 part per million (ppm): 1 part out of 1 X 106 parts 1 part per billion (ppb): 1 part out of 1 X 109 parts 1 part per trillion (ppt): 1 part out of 1 X 1012 parts 1 ppm = 1 drop in a full bathtub 1 ppb = 1 drop in a full swimming pool 1 ppt = 1 drop in 1000 full swimming pools We express ppm concentration in a variety of units depending on what we need to use.  But they are all interrelated. ppm =   1 g        =       1 g      =        1 mg       =    1 mg     =   1 microgram             106 mL        1000 L             1 L               1 kg                    1 g Example #2   Dissolved O2 in water shows a concentration of 250 mL of water At SATP and 2.2 mg of O2.  What is the concenrtaion in ppm? ppm concentration = 1 mg   =   2.2 mg   =  8.8 mg/L = 8.8 ppm                                    1 L          0.25 L Stop here and go to the Concentration Unit Calculations Other than Molarity Workshsheet

 Molar Concentrations and Molarity Molarity is a way of specifing the amount of solute in one litre of solvent.   Molarity is also known as the concentration of a solution.  The concentration of a solution is the ratio of solute to a given quantity of solvent. We can use whatever units we wish but the most commonly used units are moles of solute per litre of solution.  The special name for this ratio is the molar concentration or molarity which is abbreviated 'M'.        M =    mol solute    =        mol solute                      L soln             1000 mL soln Suppose we had a bottle with the label "0.5 M KBr".  This means that it contains a solution of potassium bromide with a concentration of 0.10 mol of potassium bromide per litre of solution (or per 1000 mL of solution).    It does not tell us the amount of solution in the bottle. Solutions are considered to be homogenous substances once the solute has completely dissolved. Sample Problem A student requires 0.250 moles of NaCl for an experiment.  The only thing available to them is a bottle with a solution labeled "0.400 M NaCl."  What volume of the solution should be used?  Give the answer in millilitres. Solution: Use the information on the bottle.  There is 0.400 moles of solute per litre.  We need 0.250 moles.                 0.400 moles = 0.400 moles = 0.250 moles                          1 L             1000 mL              x  x = 625 mL.  Thus 625 mL of 0.400 M NaCl contains 0.250 mol of NaCl. Practise Problem: A glucose solution with a molar concentration of 0.200 M is available.  What volume of this solution must be measured to obtain 0.001 mol of glucose?             Extension Exercise 2.5  Concentrations of Solutions Preparation of Solutions Another very common problem involving molar concentration is the calculation of the number of grams of solute needed to make a given volume of solution having a specific molarity.  The best way to see this is by example: Problem:  How can 500 mL of 0.150 M Na2CO3 solution be prepared? Solution: This is stated in a way that normally arises in the lab?  What we really want to know is how many grams of Na2CO3 that are going to be in 500 mL of 0.150 M Na2CO3 solution. Although the label reads in M, the balance reads in grams.  Before we can calculate the number of grams we need to know the number of moles.                    0.150 M  =  0.150 mol  = 0.150 mol  =         x                                               L            1000 mL        500 mL x = 0.075 mol of solute.  Therefore we need to weigh out 0.075 mol of Na2CO3.  The number of grams of Na2CO3 are therefore:                                            g = n * mm                                               = 0.075 mol * 105.99 g/mol                                               =  7.95 grams To answer the question:   Weigh out 7.95 grams of sodium carbonate.  Add it to 100 mL of water in a 500 mL volumetric flask.    Swirl to dissolve.  Top the flask up with 400 mL more water up to the 500 mL meniscus mark. Practise problem:  How can we prepare 250 mL of 0.200 M NaHCO3?           Stop here and go to the Molarity and Solution Creation Worksheet Homework Section 2.5 Page 137 Alternate Exercise 2.5 - House Hold product Safety Testing Tec Extension 2.6 The Breathalyzer Activity 2.7  Determining the Concentration of a Solution Alternate Activity 2.7 Concentration Using a Spectrophotometer