For this item, we find first the mass of the compound with the empirical formula. That is,
molar mass = 12 (for C) + 1 (for H)
= 13 g/mole
Then, we divide the given mass of the molecular formula by the mass of the empirical formula.
n = 52.07 / 13 = 4
Therefore, the molecular formula is C₄H₄.
Answer:
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A) Nitrogen has an ATOMIC mass number of 14, but nitrogen gas consists of N₂ molecules, so the mass to use in this problem is 28 g/mol. Rates of effusion ∝ 1/√(mass), so
<span>√(mass unknown) /√28 = (rate N₂ effusion)/(rate unknown effusion) = 1.59 </span>
<span>∴ mass unknown = (1.59)²(28) = 70.78 g/mol </span>
<span>B) One possible gas that comes close for this mass is NF₃.</span>
Density= 1.2092 because density = mass/volume