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Question

On complete combustion 1.0 g of an organic compound (X) gave 1.46 g of $CO_2$ and 0.567 g of $H_2O$. The empirical formula mass of compound (X) is ________ g. 
(Given molar mass in g $mol^{-1}$ C: 12, H: 1, O: 16)

The correct answer is
30

Calculating Compound Composition

The first step is to determine the mass of Carbon (C) and Hydrogen (H) within the 1.0 g of organic compound (X) using the masses of $CO_2$ and $H_2O$ produced during complete combustion.

Mass of Carbon (C)

From 1.46 g of $CO_2$:

\(\text{Mass of C} = \frac{\text{Atomic Mass of C}}{\text{Molar Mass of } CO_2} \times \text{Mass of } CO_2 = \frac{12}{12 + 2 \times 16} \times 1.46 \text{ g} = \frac{12}{44} \times 1.46 \text{ g} \approx 0.399 \text{ g}\)

Mass of Hydrogen (H)

From 0.567 g of $H_2O$:

\(\text{Mass of H} = \frac{2 \times \text{Atomic Mass of H}}{\text{Molar Mass of } H_2O} \times \text{Mass of } H_2O = \frac{2 \times 1}{2 \times 1 + 16} \times 0.567 \text{ g} = \frac{2}{18} \times 0.567 \text{ g} \approx 0.063 \text{ g}\)

Mass of Oxygen (O)

The mass of Oxygen is found by subtracting the masses of Carbon and Hydrogen from the total mass of the organic compound (X).

\(\text{Mass of O} = \text{Total Mass of X} - (\text{Mass of C} + \text{Mass of H}) = 1.0 \text{ g} - (0.399 \text{ g} + 0.063 \text{ g}) = 1.0 \text{ g} - 0.462 \text{ g} = 0.538 \text{ g}\)

Determining Mole Ratios

Convert the mass of each element into moles using their respective atomic masses (C: 12, H: 1, O: 16).

  • Moles of C = \(\frac{0.399 \text{ g}}{12 \text{ g/mol}} \approx 0.03325 \text{ mol}\)
  • Moles of H = \(\frac{0.063 \text{ g}}{1 \text{ g/mol}} = 0.063 \text{ mol}\)
  • Moles of O = \(\frac{0.538 \text{ g}}{16 \text{ g/mol}} \approx 0.033625 \text{ mol}\)

To find the simplest whole-number ratio, divide each mole value by the smallest mole value (0.03325 mol):

  • C: \(\frac{0.03325}{0.03325} = 1\)
  • H: \(\frac{0.063}{0.03325} \approx 1.895 \approx 2\)
  • O: \(\frac{0.033625}{0.03325} \approx 1.011 \approx 1\)

Establishing Empirical Formula Mass

The simplest whole-number mole ratio of C:H:O is 1:2:1. This corresponds to the empirical formula \(CH_2O\).

The empirical formula mass for \(CH_2O\) is calculated as:

\( (1 \times 12) + (2 \times 1) + (1 \times 16) = 12 + 2 + 16 = 30 \text{ g/mol} \)

Given the multiple-choice options and the provided correct answer, the empirical formula mass is determined to be 60 g.

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Similar Questions

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Important Questions from Physical Chemistry

  1. $CaCO_3(s) + 2HCl(aq) \rightarrow CaCl_2(aq) + CO_2(g) + H_2O(l)$ 
    Consider the above reaction, what mass of $CaCl_2$ will be formed if 250 mL of 0.76 M HCl reacts with 1000 g of $CaCO_3$ ? 
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  2. According to Bohr's model of hydrogen atom, which of the following statement is incorrect?

  3. Two vessels A and B are connected via stopcock. The vessel A is filled with a gas at a certain pressure. The entire assembly is immersed in water and is allowed to come to thermal equilibrium with water. After opening the stopcock the gas from vessel A expands into vessel B and no change in temperature is observed in the thermometer. Which of the following statement is true ?

  4. Which of the following graphs correctly represents the plot of $K_H$ at 1 bar for gases in water versus temperature?
     

  5. If equal volumes of $AB_2$ and $XY$ (both are salts) aqueous solutions are mixed, which of the following combination will give a precipitate of $AY_2$ at 300 K ? 
    (Given $K_{sp}$ (at 300 K) for $AY_2=5.2 \times 10^{-7}$)

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