Question

A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives ${\text{3}}{\text{.38 g}}$ carbon dioxide, ${\text{0}}{\text{.690 g}}$ of water and no other products. A volume of ${\text{10}}{\text{.0 L}}$ (measured at STP) of this welding gas is found to weigh ${\text{11}}{\text{.6 g}}$.
Calculate (i) empirical formula, (ii) molar mass of the gas and (iii) molecular formula.

Answer 1

Hint: The formula that gives the proportions of the elements in a compound is known as the empirical formula. The empirical formula does not give the actual number of atoms. The actual number of each atom is given by the molecular formula.

Complete step by step answer:
Step 1: Calculate the mass of carbon present in the welding fuel gas as follows:
The welding fuel gas on burning produces ${\text{3}}{\text{.38 g}}$ of carbon dioxide.
The molar mass of carbon dioxide is ${\text{44 g mo}}{{\text{l}}^{ - 1}}$ and the molar mass of carbon is ${\text{12 g mo}}{{\text{l}}^{ - 1}}$. One mole of carbon dioxide contains one mole of carbon. Thus, ${\text{3}}{\text{.38 g}}$ of carbon dioxide contains,
${\text{Mass of carbon}} = \dfrac{{12{\text{ g carbon}}}}{{{\text{44 g carbon dioxide}}}} \times {\text{3}}{\text{.38 g carbon dioxide}}$
${\text{Mass of carbon}} = {\text{0}}{\text{.92 g}}$
Thus, the mass of carbon present in the welding fuel gas is ${\text{0}}{\text{.92 g}}$.
Step 2: Calculate the mass of hydrogen present in the welding fuel gas as follows:
The welding fuel gas on burning produces ${\text{0}}{\text{.690 g}}$ of water.
The molar mass of water is ${\text{18 g mo}}{{\text{l}}^{ - 1}}$ and the molar mass of hydrogen is ${\text{1 g mo}}{{\text{l}}^{ - 1}}$. One mole of water contains two moles of hydrogen. Thus, ${\text{0}}{\text{.690 g}}$ of water contains,
${\text{Mass of hydrogen}} = \dfrac{{2 \times 1{\text{ g hydrogen}}}}{{{\text{18 g water}}}} \times {\text{0}}{\text{.690 g water}}$
${\text{Mass of hydrogen}} = {\text{0}}{\text{.077 g}}$
Thus, the mass of hydrogen present in the welding fuel gas is ${\text{0}}{\text{.077 g}}$.
Step 3: Calculate the number of moles of carbon using the equation as follows:
${\text{Number of moles}} = \dfrac{{{\text{Mass}}}}{{{\text{Molar mass}}}}$
Substitute ${\text{0}}{\text{.92 g}}$ for the mass of carbon, ${\text{12 g mo}}{{\text{l}}^{ - 1}}$ for the molar mass of carbon. Thus,
${\text{Number of moles of carbon}} = \dfrac{{{\text{0}}{\text{.92 g}}}}{{{\text{12 g mo}}{{\text{l}}^{ - 1}}}}$
${\text{Number of moles of carbon}} = 0.077{\text{ mol}}$
Thus, the number of moles of carbon are $0.077{\text{ mol}}$.
Step 4: Calculate the number of moles of hydrogen using the equation as follows:
${\text{Number of moles}} = \dfrac{{{\text{Mass}}}}{{{\text{Molar mass}}}}$
Substitute ${\text{0}}{\text{.077 g}}$ for the mass of hydrogen, ${\text{1 g mo}}{{\text{l}}^{ - 1}}$ for the molar mass of hydrogen. Thus,
${\text{Number of moles of hydrogen}} = \dfrac{{{\text{0}}{\text{.077 g}}}}{{{\text{1 g mo}}{{\text{l}}^{ - 1}}}}$
${\text{Number of moles of hydrogen}} = 0.077{\text{ mol}}$
Thus, the number of moles of hydrogen are $0.077{\text{ mol}}$.
Step 5: Determine the empirical formula of the welding fuel gas as follows:
The number of moles of carbon are $0.077{\text{ mol}}$.
The number of moles of hydrogen are $0.077{\text{ mol}}$.
The ratio of number of moles of carbon to the number of moles of hydrogen is,
${\text{C:H}} = \dfrac{{0.077{\text{ mol}}}}{{0.077{\text{ mol}}}} = {\text{1:1}}$
Thus, the ratio of the number of moles of carbon to the number of moles of hydrogen is ${\text{1:1}}$.
Thus, the empirical formula is ${\text{CH}}$.
Thus, the empirical formula of the welding fuel gas is ${\text{CH}}$.
Step 6: Determine the molar mass of the welding gas fuel as follows:
We know the Avogadro’s hypothesis which states that at standard temperature and pressure (STP), ${\text{1 mole}}$ of any gas occupies a volume of ${\text{22}}{\text{.4 L}}$.
Also, we are given that a volume of ${\text{10}}{\text{.0 L}}$ of welding gas fuel at STP is found to weigh ${\text{11}}{\text{.6 g}}$. Thus, at ${\text{22}}{\text{.4 L}}$ the weight of the gas will be its molar mass. Thus,
\[{\text{Molar mass}} = \dfrac{{11.6{\text{ g}}}}{{10.0{\text{ L}}}} \times 22.4{\text{ L}}\]
\[{\text{Molar mass}} = 26{\text{ g mo}}{{\text{l}}^{ - 1}}\]
Thus, the molar mass of the welding gas fuel is \[26{\text{ g mo}}{{\text{l}}^{ - 1}}\].
Step 7: Determine the molecular formula of the welding gas fuel as follows:
The empirical formula of the welding gas fuel is ${\text{CH}}$. Thus, the empirical formula mass of the welding gas fuel is,
${\text{Empirical formula mass}} = \left( {12 + 1} \right){\text{g mo}}{{\text{l}}^{ - 1}}$
${\text{Empirical formula mass}} = 13{\text{ g mo}}{{\text{l}}^{ - 1}}$
Thus, the empirical formula mass of the welding gas fuel is $13{\text{ g mo}}{{\text{l}}^{ - 1}}$.
The ratio of the molar mass and the empirical formula mass is known as the n-factor. Thus,
${\text{n - factor}} = \dfrac{{{\text{Molar mass}}}}{{{\text{Empirical formula mass}}}}$
${\text{n - factor}} = \dfrac{{{\text{26 g mo}}{{\text{l}}^{ - 1}}}}{{13{\text{ g mo}}{{\text{l}}^{ - 1}}}}$
${\text{n - factor}} = 2$
Thus, the n—factor is $2$.
The molecular formula is the product of n-factor and the empirical formula. Thus,
${\text{Molecular formula}} = {\text{n - factor}} \times {\text{Empirical formula}}$
${\text{Molecular formula}} = {\text{2}}\left( {{\text{CH}}} \right)$
${\text{Molecular formula}} = {{\text{C}}_{\text{2}}}{{\text{H}}_{\text{2}}}$
Thus, the molecular formula of the welding gas fuel is ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{2}}}$.
Thus, (i) Empirical formula is ${\text{CH}}$.
(ii) Molar mass of the gas is \[26{\text{ g mo}}{{\text{l}}^{ - 1}}\].
(iii) Molecular formula is ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{2}}}$.


Note:
The Avogadro’s hypothesis states that at standard temperature and pressure (STP), ${\text{1 mole}}$ of any gas occupies a volume of ${\text{22}}{\text{.4 L}}$. Thus, molar mass can be calculated using Avogadro's law.