Question

A miture of powerded aluminum and tin was burned in the atmosphere of oxygen in a way such that the resulting oxides could be collected and weighed 0.5488g; the mixture of Al2O3 and SnO2 weighed 0.7712g. Calculate the weight percent and atom percent of Al and Sn in the initial mixture

Answer 1

A miture of powered aluminum and tin was burned in the atmosphere of oxygen in a way such that the resulting oxides could be collected and weighed 0.5488g; the mixture of Al2O3 and SnO2 weighed 0.7712g. Calculate the weight percent and atom percent of Al and Sn in the initial mixture.

Solution:

1.4Al+3O₂=2Al₂O₃

Sn+O₂=SnO₂

2. Let m(Al)=x gram, so, m(Sn)=(0.5488-x)gram, according to the condition of the task.

3. M(Al)=27 gram/mole;

M(Sn)=119 gram/mole;

M(Al₂O₃)=102 gram/mole;

M(SnO₂)=151 gram/mole.

4. I make a proportion, where a, b are mass Al2O3 and SnO2:

$$\frac {x \text{ gram}}{4 \times 54 \frac {\text{gram}}{\text{mole}}} = \frac {a}{2 \times 102 \frac {\text{gram}}{\text{mole}}};$$ $$\frac {(0.5488 - x) \text{gram}}{1 \times 119 \frac {\text{gram}}{\text{mole}}} = \frac {b}{1 \times 151 \frac {\text{gram}}{\text{mole}}};$$ $$a = \frac {x \times 2 \times 102}{4 \times 27} = 1.889x;$$ $$b = \frac {(0.5488 - x) \times 151}{119} = 0.7 - 1.27x;$$

m(Al₂O₃)+m(SnO₂)=a+b=0.7712 gram;

1.889x+0.7-1.27x=0.7712

x=0.115

x=m(Al)=0.115 gram

m(Sn)=0.5488-0.115=0.4338 gram

5. $\omega(\text{Al})=\frac{0.115}{0.5488} \times 100\% = 20.95\%$

$\omega_{1}(\text{Sn})=\frac{0.4338}{0.5488} \times 100\% = 79.05\%$

6. $\text{n(Al)}=\frac{0.115}{27} = 0.0043 \, \text{mol}$

$$n(Sn) = \frac{0.4338}{119} = 0.00365 \text{ mol}$$ $$7.N(Al) = 0.0043 \times 6.02 \times 10^{23} = 0.0259 \times 10^{23}$$ $$N(Sn) = 0.00365 \times 6.02 \times 10^{23} = 0.022 \times 10^{23}$$ $$\sum \left(N(Al) + N(Sn)\right) = 0.0259 \times 10^{23} + 0.022 \times 10^{23} = 0.0479 \times 10^{23};$$ $$8.\omega_2(Al) = \frac{0.0259 \times 10^{23}}{0.0479 \times 10^{23}} \times 100\% = 54\%$$ $$\omega_3(Sn) = \frac{0.022 \times 10^{23}}{0.0479 \times 10^{23}} \times 100\% = 46\%$$

Answer: $\omega(Al) = 20.95\%$; $\omega_1(Sn) = 79.05\%$; $\omega_2(Al) = 54\%$; $\omega_3(Sn) = 46\%$.

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