A 0.5242-g sample of a compound known to contain only carbon, hydrogen, and oxygen was burned in oxygen to yield 0.9740 g of CO2 and 0.1994 g of H2O. What is the empirical formula of the compound?

C₃H₃O₂

Explanation:

From the question given above, the following data were obtained:

Mass of compound = 0.5242 g

Mass of CO₂ = 0.9740 g

Mass of H₂O = 0.1994 g

Empirical formula =?

Next, we shall determine the mass of carbon, hydrogen and oxygen present in the compound. This can be obtained as follow:

For carbon, C:

Mass of CO₂ = 0.9740 g

Molar mass of CO₂ = 12 + (16×2)

= 12 + 32 = 44 g/mol

Mass of C = 12/44 × 0.9740

Mass of C = 0.2656 g

For hydrogen, H:

Mass of H₂O = 0.1994 g

Molar mass of H₂O = (2×1) + 16

= 2 + 16 = 18 g/mol

Mass of H = 2/18 × 0.1994

Mass of H = 0.0222 g

For oxygen, O:

Mass of compound = 0.5242 g

Mass of C = 0.2656 g

Mass of H = 0.0222 g

Mass of O =?

Mass of O = (Mass of compound) – (Mass of C + Mass of H)

Mass of O = 0.5242 – (0.2656 + 0.0222)

Mass of O = 0.5242 – 0.2878

Mass of O = 0.2364 g

Finally, we shall determine the empirical formula. This can be obtained as follow:

Mass of C = 0.2656 g

Mass of H = 0.0222 g

Mass of O = 0.2364 g

Divide by their molar mass

C = 0.2656 /12 = 0.0221

H = 0.0222 /1 = 0.0222

O = 0.2364 /16 = 0.0148

Divide by the smallest

C = 0.0221 / 0.0148 = 1.5

H = 0.0222 / 0.0148 = 1.5

O = 0.0148 / 0.0148 = 1

Multiply through by 2 to express in whole number.

C = 1.5 × 2 = 3

H = 1.5 × 2 = 3

O = 1 × 2 = 2

Empirical formula => C₃H₃O₂

the density(ratio between mass and volume as m/v) decreases from top to bottom in periodic table group as the size increases and nuclear charge decreases. but in periods density increases from left to right as atom size decreases and nuclear charge increases.

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explanation:

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