A 0.4550 g mixture treated with excess Ba(NO3)2 gives 0.6168 g of BaSO4 (233.43 g/mol). What is the mass percent of MgSO4 (120.37 g/mol)?

1. Convert the precipitate to moles: 0.6168 g divided by 233.43 g/mol gives 0.0026421 mol BaSO4. 2. The equation pairs 1 mol MgSO4 with 1 mol BaSO4, so the same molar amount of MgSO4 was present. 3. Convert back to mass: 0.0026421 mol times 120.37 g/mol gives 0.3181 g MgSO4. 4. Mass percent equals the analyte mass divided by the sample mass, times 100. 5. That is 0.3181 divided by 0.4550, which is 0.699. 6. Multiplying by 100 gives about 69.9% MgSO4 in the mixture. 7. Dividing the sample mass by the precipitate mass instead gives 138%, an impossible percentage. _Source: OpenStax Chemistry (CC BY 4.0), Ch 4 "Stoichiometry of Chemical Reactions", section 4.5 Quantitative Chemical Analysis_