Worksheet with chemistry practice problems on molar mass calculations, including formula masses, molar masses, mass, and number of moles for different compounds.

a) H₃PO₄: 3(1.008) + 30.97 + 4(16.00) = 97.99 ≈ 98 amu
b) C₄H₁₀: 4(12.01) + 10(1.008) = 58.12 ≈ 58 amu
c) Co(NO₃)₂: 58.93 + 2(14.01) + 6(16.00) = 182.95 ≈ 183 amu
d) FeCl₃: 55.85 + 3(35.45) = 162.20 ≈ 161 amu
e) C₁₂H₂₂O₁₁: 12(12.01) + 22(1.008) + 11(16.00) = 342.30 ≈ 342 amu
f) Al₂(SO₄)₃: 2(26.98) + 3(32.07) + 12(16.00) = 342.15 ≈ 342 amu

a) H₂O: 2(1.008) + 16.00 = 18.016 ≈ 18 g/mol
b) H₂SO₄: 2(1.008) + 32.07 + 4(16.00) = 98.086 ≈ 98 g/mol
c) C₆H₁₂O₆: 6(12.01) + 12(1.008) + 6(16.00) = 180.156 ≈ 180 g/mol
d) C₇H₈N₄O₂: 7(12.01) + 8(1.008) + 4(14.01) + 2(16.00) = 180.166 ≈ 180 g/mol
e) (Na)C₁₁H₁₈N₂O₃: 22.99 + 11(12.01) + 18(1.008) + 2(14.01) + 3(16.00) = 249.28 ≈ 249 g/mol
f) (C₅H₅N)₄Cu(NCS)₆: 4(65.09) + 63.55 + 6(12.01 + 14.01 + 32.07) = 728.32 ≈ 728 g/mol

a) 1.00 mol NaCl: 1.00 × (22.99 + 35.45) = 58.44 g ≈ 58 g
b) 2.00 mol MgBr₂: 2.00 × (24.31 + 2(79.90)) = 368.22 g ≈ 368 g
c) 0.50 mol CCl₄: 0.50 × (12.01 + 4(35.45)) = 76.01 g ≈ 76 g
d) 3.60 mol NaOH: 3.60 × (22.99 + 16.00 + 1.008) = 144.03 g ≈ 144 g
e) 0.710 mol Ca(CN)₂: 0.710 × (40.08 + 2(12.01 + 14.01)) = 65.32 g
f) 1.82 mol BaSO₄: 1.82 × (137.33 + 32.07 + 4(16.00)) = 424.06 g

a) 88 g CO₂: 88 / (12.01 + 2(16.00)) = 88 / 44.01 = 2.00 moles
b) 198 g C₂H₁₂O₆: 198 / (2(12.01) + 12(1.008) + 6(16.00)) = 198 / 180.16 = 1.5 moles
c) 65.25 g LiBr: 65.25 / (6.94 + 79.90) = 65.25 / 86.84 = 0.75 moles
a) 249 g NaOH: 249 / (22.99 + 16.00 + 1.008) = 249 / 40.00 = 6.225 moles
b) 878 g NaAlSi₃O₈: 878 / (22.99 + 26.98 + 3(28.09) + 8(16.00)) = 878 / 262.14 = 3.35 moles
c) 7,327 g Cu₂FeSnS₄: 7327 / (2(63.55) + 55.85 + 118.71 + 4(32.07)) = 7327 / 431.02 = 17.00 moles