Question

An oxygen enriched air, which contains at least 50% by volume pure oxygen, has been sold by a company. To test their claim, a 1000 mL glass bulb was filled with the gas to a total pressure of 750 mm Hg at 68°F. The weight of the gas sample was found to be 1.236 g. What is the exact volume percent of oxygen gas in the mixture?

Answer 1

Answer on Question #56881 – Chemistry - General Chemistry

Question:

An oxygen enriched air, which contains at least 50% by volume pure oxygen, has been sold by a company. To test their claim, a 1000 mL glass bulb was filled with the gas to a total pressure of 750 mm Hg at 68°F. The weight of the gas sample was found to be 1.236 g. What is the exact volume percent of oxygen gas in the mixture?

Answer:

$$68^{\circ}F = 293.15 \, \text{K}$$ $$R = 62.364 \, (\text{mm} \cdot \text{l}) / (\text{mol} \cdot \text{K})$$

We will assume the 50% of the volume is occupied by O₂ and other 50% - by N₂.

$$M(O_2) = 32 \, \text{g/mol}$$ $$M(N_2) = 28 \, \text{g/mol}$$ $$\begin{array}{l} pV = vRT \\ 750 \cdot 0.5 = \frac{m}{32} \cdot 62.364 \cdot 293.15 \\ m(O_2) = 0.6564 \, \text{g} \end{array}$$ $$\begin{array}{l} 750 \cdot 0.5 = \frac{m}{28} \cdot 62.364 \cdot 293.15 \\ m(N_2) = 0.5743 \, \text{g} \end{array}$$ $$m = 0.6564 + 0.5743 = 1.2307$$

If 50%-Oxygen air weigh 1.236 g, than the amount of Oxygen in the current sample is:

$$(1.237 \times 50) / 1.236 = 50.215\%$$

Which is very close to the started value.

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