U1.1 · Moles and Molar MassPractice · 摩尔与摩尔质量

Moles of atoms = (1.00 g ÷ M) × atoms per formula. Per gram: $\ce{He}$: 1 ÷ 4.00 = 0.250 mol; $\ce{H2O}$: 3 ÷ 18.02 = 0.166 mol; $\ce{CH4}$: 5 ÷ 16.04 = 0.312 mol; $\ce{NH3}$: 4 ÷ 17.03 = 0.235 mol. $\ce{CH4}$ has a small molar mass and 5 atoms per molecule, so it has the most atoms. Picking He just for its low molar mass misses that.

Each $\ce{Ca(NO3)2}$ has 2 $\ce{NO3-}$ ions, so 2 × 3 = 6 O. Moles of O = 0.250 × 6 = 1.50 mol, and the number of atoms = 1.50 × $6.022 \times 10^{23}$ = $9.03 \times 10^{23}$. The subscript 2 outside the parentheses applies to the whole $\ce{NO3}$; counting only 3 O gives half the answer.

Same number of molecules means the same number of moles. 16.0 g of $\ce{O2}$ is 16.0 ÷ 32.00 = 0.500 mol. The choices: 16.0 g $\ce{CH4}$ = 0.998 mol, 22.0 g $\ce{CO2}$ = 0.500 mol, 18.0 g $\ce{H2O}$ = 0.999 mol, 4.0 g $\ce{H2}$ = 1.98 mol. Equal masses don’t mean equal numbers of molecules.

Molar mass = mass ÷ moles = 1.60 ÷ 0.0250 = 64.0 g/mol. Molar masses (g/mol): $\ce{O2}$ 32.00, $\ce{CO2}$ 44.01, $\ce{SO2}$ 64.06, $\ce{Cl2}$ 70.90. The closest is $\ce{SO2}$.

n = $1.00 \times 10^{22}$ ÷ $6.022 \times 10^{23}$ = 0.01661 mol, and mass = 0.01661 mol × 196.97 g/mol = 3.27 g. Particles → moles → grams: always go through moles.