Percentage Purity Of A Sample
We can use the HCl standardized solution from above to analyze an impure sample of Na2CO3. In a calculation of this type of analysis, convert the moles of unknown into mass units and then apply the general formula:
 
% of component = grams of component * 100
                                     grams of sample
What is the percentage of Na2CO3 in an impure sample if 25.00 mL of 0.315 mol/L HCl is required to react completely with a 0.600 gram sample of the impure salt? The impurities do not react with HCl.
 
2 HCl(aq) + Na2CO3 ------> 2 NaCl(aq) + CO2(g) + H2O(l)

 
moles of HCl  =  0.315 mol/L * 0.025 L = 0.007875 moles of HCl
 
0.007875 mol of HCl * 1 mole of Na2CO3 = 0.00394 moles Na2CO3
                                            2 moles HCl
 
0.00394 moles of Na2CO3 * 106.00 grams/moles = 0.418 g Na2CO3
 
% purity = 0.418 g * 100% = 69.7%
                   0.600 g
 
Follow-Up Problem
What is the percentage of acetic acid, CH3COOH, in a sample of vinegar if 35.00 mL of 0.468 M NaOH solution is required to neutralize a 25.00 mL sample of the vinegar which has a density of 1.06 g/mL. Acetic acid is monoprotic and has a mm = 60.00 g/mole.