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 litre (measured at STP) of this welding gas is found to weigh 11.6 g. Calculate molecular formula of gas.
Question
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 litre (measured at STP) of this welding gas is found to weigh 11.6 g. Calculate molecular formula of gas.
Solution
Step 1: Determine the moles of carbon and hydrogen in the sample
From the given data, we know that the combustion of the gas produced 3.38 g of carbon dioxide (CO2) and 0.690 g of water (H2O).
The molar mass of CO2 is approximately 44.01 g/mol, and the molar mass of H2O is approximately 18.02 g/mol.
Therefore, the number of moles of CO2 is 3.38 g / 44.01 g/mol = 0.0768 mol, and the number of moles of H2O is 0.690 g / 18.02 g/mol = 0.0383 mol.
Since each mole of CO2 contains one mole of carbon, there are 0.0768 mol of carbon in the sample.
Each mole of H2O contains two moles of hydrogen, so there are 0.0383 mol * 2 = 0.0766 mol of hydrogen in the sample.
Step 2: Determine the empirical formula of the gas
The empirical formula is the simplest, most reduced ratio of elements in a compound.
Since the moles of carbon and hydrogen are approximately equal, the empirical formula of the gas is CH.
Step 3: Determine the molar mass of the gas
The given data tells us that 10.0 L of the gas at STP weighs 11.6 g.
We know that one mole of any gas at STP occupies a volume of 22.4 L. Therefore, the number of moles in 10.0 L of the gas is 10.0 L / 22.4 L/mol = 0.446 mol.
The molar mass of the gas is the mass of the sample divided by the number of moles, or 11.6 g / 0.446 mol = 26 g/mol.
Step 4: Determine the molecular formula of the gas
The molecular formula is a multiple of the empirical formula. The molar mass of the empirical formula CH is approximately 12.01 g/mol for carbon + 1.008 g/mol for hydrogen = 13.02 g/mol.
The ratio of the molar mass of the gas to the molar mass of the empirical formula is 26 g/mol / 13.02 g/mol = 2.
Therefore, the molecular formula of the gas is C2H2.
Similar Questions
Acetylene C2H2 gas is often used in welding torches because of the very high heat produced when it reacts with oxygen O2 gas, producing carbon dioxide gas and water vapor. Calculate the moles of water produced by the reaction of 0.095mol of oxygen. Be sure your answer has a unit symbol, if necessary, and round it to 2 significant digits.
A 2.554-g sample of a certain hydrocarbon is burned in excess oxygen, producing 8.635 g CO2(g) and 1.768 g H2O(l). If the molecular weight of the hydrocarbon is found to be 78.11 u, what is its molecular formula? [m.w. CO2 = 44.01 u; m.w. H2O = 18.02 u]
____ is a gas that is produced by the ultraviolet radiation in the air in the vicinity of welding operations.Group of answer choicesArgonPhosgeneOzoneCarbon dioxide
A major component of gasoline is octane C8H18. When liquid octane is burned in air it reacts with oxygen O2 gas to produce carbon dioxide gas and water vapor. Calculate the moles of oxygen needed to produce 1.20mol of water. Be sure your answer has a unit symbol, if necessary, and round it to 3 significant digits.
Acetylene (C2H2) burns in oxygen (O2) to produce carbon dioxide (CO2) and water (H2O). 8.64 L of gaseous acetylene (C2H2) is burned at STP according to the equation: 2C2H2 (g) + 5O2 (g) ⟶ 4CO2 (g) + 2H2O (l)a. What is the volume of O2 required for complete combustion?b. What is the volume of CO2 produced upon complete combustion?c. How many moles of CO2 are produced upon complete combustion?
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.