Chemistry / Unit 8: Stoichiometry
8.2 Mass-to-Mass Stoichiometry
Notes
To go from grams of one substance to grams of another, use three steps: grams A $\rightarrow$ moles A $\rightarrow$ moles B $\rightarrow$ grams B.
- Step 1: divide by the molar mass of A.
- Step 2: multiply by the mole ratio from the balanced equation.
- Step 3: multiply by the molar mass of B.
- Example: $2\text{Na} + \text{Cl}_2 \rightarrow 2\text{NaCl}$. From 46.0 g Na: $\displaystyle \frac{46.0}{22.99} = 2.00$ mol Na $\rightarrow$ 2.00 mol NaCl $\rightarrow 2.00 \times 58.44 = 117$ g NaCl.
Practice
In $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$, what mass of water forms from 10.0 g of $\text{H}_2$? ($\text{H}_2$ = 2.02 g/mol, $\text{H}_2\text{O}$ = 18.02 g/mol)
$\displaystyle \frac{10.0}{2.02} = 4.95$ mol $\text{H}_2$ $\rightarrow$ 4.95 mol $\text{H}_2\text{O}$ (1 : 1), and $4.95 \times 18.02 = 89.2$ g.
In $\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$, what mass of $\text{CO}_2$ forms from 50.0 g of $\text{CaCO}_3$? ($\text{CaCO}_3$ = 100.09 g/mol, $\text{CO}_2$ = 44.01 g/mol)
$\displaystyle \frac{50.0}{100.09} = 0.500$ mol $\text{CaCO}_3$ $\rightarrow$ 0.500 mol $\text{CO}_2$, and $0.500 \times 44.01 = 22.0$ g.
In $2\text{KClO}_3 \rightarrow 2\text{KCl} + 3\text{O}_2$, what mass of $\text{O}_2$ forms from 24.5 g of $\text{KClO}_3$? ($\text{KClO}_3$ = 122.55 g/mol, $\text{O}_2$ = 32.00 g/mol)
$\displaystyle \frac{24.5}{122.55} = 0.200$ mol $\text{KClO}_3$. $\displaystyle 0.200 \times \frac{3}{2} = 0.300$ mol $\text{O}_2$, and $0.300 \times 32.00 = 9.60$ g. 6.40 g skips the mole ratio.
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