Question

A 0.580 g sample of a compound containing only carbon and hydrogen contains 0.480 g of carbon and 0.100 g of hydrogen. At STP, 33.6 mL of the gas has a mass of 0.087 g. What is the molecular (true) formula for the compound

Answer 1

Answer:

Molecular formula for the gas is: C₄H₁₀

Explanation:

Let's propose the Ideal Gases Law to determine the moles of gas, that contains 0.087 g

At STP → 1 atm and 273.15K

1 atm . 0.0336 L = n . 0.082 . 273.15 K

n = (1 atm . 0.0336 L) / (0.082 . 273.15 K)

n = 1.500 × 10⁻³ moles

Molar mass of gas = 0.087 g / 1.500 × 10⁻³ moles = 58 g/m

Now we propose rules of three:

If 0.580 g of gas has ____ 0.480 g of C _____ 0.100 g of C

58 g of gas (1mol) would have:

(58 g . 0.480) / 0.580 = 48 g of C  

(58 g . 0.100) / 0.580 = 10 g of H

 48 g of C / 12 g/mol = 4 mol

 10 g of H / 1g/mol = 10 moles

Answer 2

The molecular formula of the compound is C4H10.

At STP;

P = 1 atm

T = 273 K

V = 33.6 mL or 0.0336 L

R = 0.082 atmLK-1mol-1

n = ?

Hence;

n = PV/RT

n = 1 atm × 0.0336 L/0.082 atmLK-1mol-1 × 273 K

n = 0.0015 moles

Number of moles = mass/molar mass

Molar mass= Mass/Number of moles

Molar mass = 0.087 g/0.0015 moles

Molar mass = 58 g/mol

Mass of carbon = (58 g × 0.480) / 0.580 = 48 g of C  

Mass of hydrogen = (58 g × 0.100) / 0.580 = 10 g of H

Number of moles of carbon = 48 g of C / 12 g/mol = 4 mol

Number of moles of hydrogen = 10 g of H / 1g/mol = 10 moles

Formula of the compound must then be C4H10.

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