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500 ml of gaseous hydrocarbon when burnt in excess of O2 gaves 2.5 Litre of CO2 and 3 litre of water vapours under same condition molecular formula of the hydrocarbon is C4 h8 C5 h10 C4 h10 C5 h12?
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500 ml of gaseous hydrocarbon when burnt in excess of O2 gaves 2.5 Lit...
C5h12okay, so you have to use principle of atom conservation, which states no. of moles of an atom of the reactant=no.of moles of atoms in the product. Let's assume the hydrocarbon as CrHy ( there are r atoms of c and h atoms of y) the reaction will be CrHy + O2= CO2 +H2OFirst, I'm gonna conserve the no. of c atoms. r x (0.5/22.4) = 1 x (2.5/22.4) since, co2 has 1 carbon. you get r=5now, conserve h atoms, y x ( 0.5/22.4) = 2 x (3/22.4) as h20 has 2 hydrogen I'm multiplying by 2.you'll get y =12so the hydrocarbon is c5y12
Community Answer
500 ml of gaseous hydrocarbon when burnt in excess of O2 gaves 2.5 Lit...
Given Information:
To Find: Molecular formula of the hydrocarbon.
Explanation:
Step 1: Calculate the number of moles of CO2 and water vapour produced.
Using the Ideal Gas Law, PV = nRT, we can calculate the number of moles of CO2 and water vapour produced.
Step 2: Calculate the number of moles of carbon and hydrogen in the hydrocarbon.
Step 3: Determine the empirical formula of the hydrocarbon.
Step 4: Determine the molecular formula of the hydrocarbon.
- Volume of gaseous hydrocarbon = 500 ml
- Volume of CO2 produced = 2.5 L
- Volume of water vapour produced = 3 L
To Find: Molecular formula of the hydrocarbon.
Explanation:
Step 1: Calculate the number of moles of CO2 and water vapour produced.
- Volume of CO2 = 2.5 L
- Volume of water vapour = 3 L
- Standard temperature and pressure (STP) = 0°C and 1 atm pressure
Using the Ideal Gas Law, PV = nRT, we can calculate the number of moles of CO2 and water vapour produced.
- For CO2: n(CO2) = PV/RT = (1 atm x 2.5 L) / (0.0821 L atm/mol K x 273 K) = 0.103 moles
- For water vapour: n(H2O) = PV/RT = (1 atm x 3 L) / (0.0821 L atm/mol K x 273 K) = 0.123 moles
Step 2: Calculate the number of moles of carbon and hydrogen in the hydrocarbon.
- For carbon: Since each mole of CO2 contains 1 mole of carbon, the number of moles of carbon in the hydrocarbon = n(CO2) = 0.103 moles
- For hydrogen: Since each mole of water vapour contains 2 moles of hydrogen, the number of moles of hydrogen in the hydrocarbon = 2 x n(H2O) = 0.246 moles
Step 3: Determine the empirical formula of the hydrocarbon.
- Divide the number of moles of carbon and hydrogen by the smallest number of moles to get the ratio of carbon to hydrogen atoms: C = 0.103/0.103 = 1, H = 0.246/0.103 = 2.38 (approximately)
- Round off the ratio of H/C to the nearest whole number: H/C = 2.38 ≈ 2
- The empirical formula of the hydrocarbon is therefore CH2.
Step 4: Determine the molecular formula of the hydrocarbon.
- The empirical formula mass of CH2 is 14 g/mol (12 g/mol for carbon + 2 g/mol for hydrogen).
- The molar mass of the hydrocarbon can be determined by dividing the mass of 500 ml of the gas (which is equal to its density) by its molar volume at STP (22.4 L/mol): Molar mass = (500/1000 g) / (22.4 L/mol) = 0.0223 g/mol
- The molecular formula mass is a multiple of the empirical formula mass: Molecular formula mass = n x empirical formula mass = n x 14
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500 ml of gaseous hydrocarbon when burnt in excess of O2 gaves 2.5 Litre of CO2 and 3 litre of water vapours under same condition molecular formula of the hydrocarbon is C4 h8 C5 h10 C4 h10 C5 h12?
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500 ml of gaseous hydrocarbon when burnt in excess of O2 gaves 2.5 Litre of CO2 and 3 litre of water vapours under same condition molecular formula of the hydrocarbon is C4 h8 C5 h10 C4 h10 C5 h12? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about 500 ml of gaseous hydrocarbon when burnt in excess of O2 gaves 2.5 Litre of CO2 and 3 litre of water vapours under same condition molecular formula of the hydrocarbon is C4 h8 C5 h10 C4 h10 C5 h12? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 500 ml of gaseous hydrocarbon when burnt in excess of O2 gaves 2.5 Litre of CO2 and 3 litre of water vapours under same condition molecular formula of the hydrocarbon is C4 h8 C5 h10 C4 h10 C5 h12?.
500 ml of gaseous hydrocarbon when burnt in excess of O2 gaves 2.5 Litre of CO2 and 3 litre of water vapours under same condition molecular formula of the hydrocarbon is C4 h8 C5 h10 C4 h10 C5 h12? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about 500 ml of gaseous hydrocarbon when burnt in excess of O2 gaves 2.5 Litre of CO2 and 3 litre of water vapours under same condition molecular formula of the hydrocarbon is C4 h8 C5 h10 C4 h10 C5 h12? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 500 ml of gaseous hydrocarbon when burnt in excess of O2 gaves 2.5 Litre of CO2 and 3 litre of water vapours under same condition molecular formula of the hydrocarbon is C4 h8 C5 h10 C4 h10 C5 h12?.
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