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.