1.34. A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 L (measured at S.T.P.) of this welding gas is found to weigh 11.6 g. Calculate (i) empirical formula, (ii) molar mass of the gas, and (iii) molecular formula.

<p><strong>1.34.&nbsp; (i)</strong> 1 mole (44 g) of CO₂ will have 12 g carbon.</p><p>So, 3.38 g of CO₂ will have carbon = 12g/44g * 3.38</p><p>= 0.9217 g</p><p>18 g of water will have 2 g of hydrogen.</p><p>So, 0.690 g of water contain hydrogen = 2g/18g&nbsp;*&nbsp;0.6902g</p><p>= 0.0767 g</p><p>Since carbon and hydrogen are the only constituents of the compound, the total mass of the compound is:</p><p>= 0.9217 g + 0.0767 g</p><p>=0.9984 g</p><p>So, the percentage of Carbon in the compound = 0.9217/0.9984&nbsp;*&nbsp;100 = 92.32%</p><p>Now, percentage of Hydrogen in the compound = 0.0767/0.9984&nbsp;*&nbsp;100 = 7.68%</p><p>Moles of carbon in the compound = 92.32/12=7.69</p><p>Moles of hydrogen in the compound = 7.68/1=7.68</p><p>Since, we have the number of moles of both the elements, the ratio of carbon to hydrogen will be:</p><p>7.69: 7.68 = 1: 1&nbsp;</p><p>This is in the whole number, so the empirical formula will be CH.</p><p><strong> (ii) </strong>Molar mass of the gas</p><p><strong>Ans: </strong>We are given,</p><p>Weight of 10.0 L of the gas (at S.T.P) = 11.6 g</p><p>Since, 1mol of a gas occupies 22.4L volume at STP. So,</p><p>Weight of 22.4L of gas at STP =&nbsp;11.6g/10.0L&nbsp;*&nbsp;22.4L</p><p>=25.984g</p><p>≈26g</p><p>Hence, the molar mass of the gas is 26 g/mol.</p><p><strong> (iii) </strong>Molecular formula</p><p>Empirical formula mass is CH = 12 + 1= 13 g</p><p>No of moles, n = Molar mass of gas/ Empirical formula mass of gas</p><p>= 26/13 = 2</p><p>Therefore, molecular formula = 2 x CH = C₂H₂</p>

0 Upvotes 0 Upvotes 0 Downvotes

• • Report Abuse
• Comment