Stoichiometric Calculations: Gram to Gram Calculations
The type of problems involved here can be solved easily with 2-3 simple steps.  Use the Bridge Method for solving only after the equation is balanced. 

Never, ever assume that an equation is balanced.
 
Example Solution #1
During its combustion, ethane C2H6, combines with oxygen O2 to give carbon dioxide and water.  A sample of ethane was burned completely and the water that formed has a mass of 1.61 grams.  How much ethane, in moles and in grams, was in the sample?

Solution:
1.  Set up the equation based on the words in the problem.  Then balance it correctly.

                              2  C2H6    +  7 O2      -------->  4  CO2   +  6  H2O
 

2.  Compute the needed formula molecular masses.   We need the molecular mass for water and for      ethane.    H2O = 18.02 g.mol;   C2H6 = 30.08 g/mol
 
3.   Draw a roadmap outline the math steps that need to be taken:

                  moles of water  --------->  moles of ethane
                           ^                                        |
                           |                                        V
                 grams of water                   grams of ethane
 


Therefore this type of question requires a three step solution.
     Step #1  Convert 1.61 grams of water to moles of water.
     Step #2  Using the equation, compare the moles of water made from moles of ethane.
     Step #3  Convert the moles of ethane back into grams of ethane.
 
Draw a roadmap outlining the mathematical steps that need to be taken:

                  moles of water  --------->  moles of ethane
                            ^                                        |
                            |                                        V
                  grams of water                  grams of ethane
 

Math Step #1  Convert 1.61 grams of water into moles of water
                  Moles = g/mm = 1.61 g/ 18.02 g/mol = 0.09 mol of water were used.
 

Math Step #2   The chemical equation shows us that 2 moles of C2H6 is needed to make 6 moles of water.  Setup a ratio:

                                          6 mol H2O = 2 mol C2H6
                                              0.09 mol            x               

 x = 0.03 moles of ethane are needed.

 


Math Step #3   Convert the moles of ethane back into grams of ethane.
 
 g = n * mm = 0.03 mol * 30.08 g/mol = 0.90 grams of ethane

Finish off the question with a statement.  1.61 grams of water can be made from 0.03 moles of ethane or 0.90 grams of ethane.


Example Question #2
Calculate how many grams of K2Cr2O7 are needed to make 35.8 grams of I2 according to the following equation.

            K2Cr2O7  +  6 NaI  +  7 H2SO4   --->
                                      Cr2(SO4)3 + 3 I2 + 7 H2O + 3 Na2SO4 + K2SO4

The equation is already balanced but you should check it over just to be sure.

Next calculate the molecular masses that you need. The question deals with iodine, I2, and potassium dichromate, K2Cr2O7.  Calculate the molecular masses for these only.  The other parts of the equation can be forgotten about.

                    I2 =  253.82 g/mol      K2Cr2O7 = 294.20 g/mol


Draw a roadmap outlining the mathematical steps that need to be taken:

                  moles of iodine  --------->  moles of potassium dichromate
                           ^                                        |
                           |                                        V
                 grams of iodine                  grams of potassium dichromate


Math Step #1  Convert grams of iodine into moles of iodine.

               moles = g/mm = 35.8 g / 253.82 g/mol = 0.14 moles of iodine
 


Math Step #2  Compare using the equation coefficients, the number of moles of potassium dichromate is need to make how many moles of iodine.

                 3 I2    =   1 K2Cr2O7
              0.14 mol          x                             

x = 0.05 moles of potassium dichromate
 


Math Step #3 Convert the moles of potassium dichromate back into grams of potassium dichromate.

       g = n * mm = 0.05 mol *  294.20 g/mol = 14.71 grams

To make 35.8 grams of iodine you must start with 14.71 grams of potassium dichromate.
 

          Stoichiometry Gram to Gram Calculations Worksheet
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