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puteri [66]
3 years ago
5

A 8.00g of a certain Compound X, known to be made of carbon, hydrogen and perhaps oxygen, and to have a molecular molar mass of

44./gmol, is burned completely in excess oxygen, and the mass of the products carefully measured: product mass carbon dioxide 24.01g water 13.10g Use this information to find the molecular formula of X.
Chemistry
1 answer:
Ne4ueva [31]3 years ago
4 0

Answer:

C3H6.

Explanation:

Data obtained from the question:

Mass of the compound = 8g

Mass of CO2 = 24.01g

Mass of H2O = 13.10g

Next, we shall determine the mass of C, H and O present in the compound. This is illustrated below:

Molar Mass of CO2 = 12 + (2x16) = 44g/mol

Molar Mass of H2O = (2x1) + 16 = 18g/mol

Mass of C in compound = Mass of C/Molar Mass of CO2 x 24.01

=> 12/44 x 24.01 = 6.5g

Mass of H in the compound = Mass of H/Molar Mass of H2O x 13.1

=> 2x1/18 x 13.1 = 1.5g

Mass of O in the compound = Mass of compound – (mass of C + Mass of H)

=> 8 – (6.5 + 1.5) = 0

Next, we shall determine the empirical formula of the compound. This is illustrated below:

C = 6.5g

H = 1.

Divide by their molar mass

C = 6.5/12 = 0.54

H = 1.4/1 = 1.

Divide by the smallest

C = 0.54/0.54 = 1

H = 1/0.54 = 2

Therefore, the empirical formula is CH2

Finally, we shall determine the molecular formula as follow:

The molecular formula of a compound is a multiple of the empirical formula.

Molecular formula = [CH2]n

[CH2]n = 44

[12 + (2x1)]n = 44

14n = 44

Divide both side by 14

n = 44/14

n = 3

Molecular formula = [CH2]n = [CH2]3 = C3H6

Therefore, the molecular formula of the compound is C3H6

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Answer:

Part 1: C₉H₂₀ (l) + 14O₂ (g) ----> 9CO₂ (g) + 10H₂0 (g)

Part 2: Volume of CO₂ produced = 1223.21 L

<em>Note: the complete second part of the question is given below:</em>

<em>2. Suppose 0.470 kg of nonane are burned in air at a pressure of exactly 1 atm and a temperature of 17.0 °C. Calculate the volume of carbon dioxide gas that is produced. Round your answer to 3 significant digits.</em>

Explanation:

Part 1: Balanced chemical equation

C₉H₂₀ (l) + 14O₂ (g) ----> 9CO₂ (g) + 10H₂0 (g)

Part 2: volume of carbon dioxide produced

From the equation of the reaction;

At s.t.p., I mole of  C₉H₂₀ reacts with 14 moles of O₂ to produce 9 moles of CO₂

molar mass of  C₉H₂₀  = 128g/mol: molar mass of CO₂ = 44 g/mol, molar volume of gas at s.t.p. = 22.4 L

Therefore, 128 g of C₉H₂₀ produces 14 * 22.4 L of CO₂ i.e. 313.6 L of CO₂.

O.470 Kg  of nonane = 470 g of nonane

470 g of C₉H₂₀ will produce 470 * (313.6/128) L of CO₂ = 1151.50 L of CO₂

Volume of CO₂ gas produced at 1 atm and 17 °C;

Using P₁V₁/T₁ = P₂V₂/T₂

V₂ = P₁V₁T₂/P₂T₁

where P₁ = 1 atm, V₁ = 1151.50 L, T₁ = 273 K, P₂ = 1 atm, T₂ = 17 + 273 = 290 K

Substituting the values; V₂ = (1 * 1151.5 * 290)/(1 * 273)

Therefore volume of CO₂ produced, V₂ = 1223.21 L of CO₂

3 0
3 years ago
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