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Empirical formula: Combustion reaction

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Wize University Chemistry Textbook > Stoichiometry

Combustion Reactions + Empirical and Molecular Formulae

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When 2.11 g of an unknown hydrocarbon (compound made up of only carbon and hydrogen atoms) was burned in excess oxygen as part of a combustion analysis. In the process 6.32 g of CO2 and 3.45 g of H2O are produced. What is the mass percent of carbon? What is the empirical formula of the hydrocarbon?


More Combustion Reactions + Empirical and Molecular Formulae Questions:
Empirical Formula
A 1.00 g sample of a species containing only C, H, and O was completely combusted in pure oxygen, and 1.5 g CO2 and 0.41g H2O were obtained. What is the empirical formula of the compound?
Combustion and Empirical Formula
A hydrocarbon (0.875 g) containing only C, H, and O yielded 2.21 g CO2 and 0.387 g H2O after complete combustion.
What is the empirical formula?
Combustion, Stoichiometry
What mass of KClO3 is required to produce proper amount of O2 needed to achieve complete combustion of 78.88 g CH4?
2 KClO3 → 2 KCl + 3 O2
Combustion Analysis Calculation
When 2.11 g of an unknown hydrocarbon (compound made up of only carbon and hydrogen atoms) was burned in excess oxygen as part of a combustion analysis. In the process 6.32 g of CO2 and 3.45 g of H2O are produced. What is the empirical formula of the hydrocarbon?
Stoichiometry 2
An unknown hydrocarbon contains 40.0% C and 6.66% H by weight.
a. What is the empirical formula of the compound?
b. What is the molecular formula if molecular weight of the compound is 180 gmol-1?
Gas stoichiometry and Empirical Formula
Complete combustion of a hydrocarbon yields 1.5 L CO2 and 2 L H2O at 298 K and 101 kPa. Whats the empirical formula of the compound?
Empirical Formula
A 1.00 g sample of a species containing only C, H, and O was completely combusted in pure oxygen, and 1.5 g CO2 and 0.41g H2O were obtained. What is the empirical formula of the compound? (enter answer: C_H_O_ For example C2H2O5)
Empirical Formulae
This question focuses on material from 1.3. Empirical Formulae.
Empirical Formulae
3.0 g of an organic compound, CxHyOz, is combusted in air to produce 7.75 g of carbon dioxide and 3.70 g of water vapour. Determine the empirical formula of this compound.
Empirical Formulae
An organic compound, CxHyClz, was subject to elemental analysis, and the compound was found to contain 64.6% carbon by mass and 11.5% hydrogen by mass. Determine the empirical formula of this compound.
Mass Percentage Practice
0.2976g of an unknown compound gives 0.3147g of CO2 and 0.3026g of H2O on combustion. The only other element known to be present is N. What is the empirical formula?
Empirical Formulae
5. 5.0 g of an organic compound, CxHyOz, is combusted in air to produce 11.87 g of carbon dioxide and 6.08 g of water vapour. Determine the empirical formula of this compound.
Empirical formula: Combustion reaction
When 2.11 g of an unknown hydrocarbon (compound made up of only carbon and hydrogen atoms) was burned in excess oxygen as part of a combustion analysis. In the process 6.32 g of CO2 and 3.45 g of H2O are produced. What is the empirical formula of the hydrocarbon?
Combustion Analysis
For a hydrocarbon (a molecule made up of only C and H atoms) combustion analysis allows us to determine the: