CH4C2H3C3H12C4H16IIIVChemical analysis of a gaseous compound showed that it contained 75.0 percent carbon (C)and 25.0 percent hydrogen (H) by mass.At 200°C, 97.2 L of the compound exerted a pressure of 0.74 atm.If the mass of 97.2 L of the compound was 59.2 g, calculate the molecular formula of thecompound. (Molar mass of C is 12 g/mole, molar mass of H is 1 g/mole) The average atomic mass of hypothetical element 71X(70.9247 amu) and 67X (66.9256 amu) is 69.72 amu.Calculate the natural abundances of the X isotopes.A- 71X 60.3 %67X 39.7 %В- 71X 39.7 %67X 60.3 %С-71х 29.7%67X 70.3 %D- 71X 69.9 %67X 30.1 %

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ISBN:9781337398909

Author:Lawrence S. Brown, Tom Holme

Publisher:Lawrence S. Brown, Tom Holme

Chapter2: Atoms And Molecules

Section: Chapter Questions

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Concept explainers

Stoichiometry

Stoichiometry is a chemical division that evaluates the quantitative content of the chemical process with the help of the reacting molecules or products taking part in the reaction. The term “stoicheion” in “Stoichiometry”, is an old Greek term for "element" whereas “metron” means "measure".

Question

CH4
C2H3
C3H12
C4H16
II
IV
Chemical analysis of a gaseous compound showed that it contained 75.0 percent carbon (C)
and 25.0 percent hydrogen (H) by mass.
At 200°C, 97.2 L of the compound exerted a pressure of 0.74 atm.
If the mass of 97.2 L of the compound was 59.2 g, calculate the molecular formula of the
compound. (Molar mass of C is 12 g/mole, molar mass of H is 1 g/mole)

The average atomic mass of hypothetical element 71X
(70.9247 amu) and 67X (66.9256 amu) is 69.72 amu.
Calculate the natural abundances of the X isotopes.
A- 71X 60.3 %
67X 39.7 %
В- 71X 39.7 %
67X 60.3 %
С-71х 29.7%
67X 70.3 %
D- 71X 69.9 %
67X 30.1 %

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