Math Problem Statement

A 2.77 g sample of a compound containing only carbon, hydrogen, and oxygen burns in excess oxygen to produce 4.06 g of CO2 and 1.66 g of H20. Calculate the mass of oxygen in the sample in g

Solution

To calculate the mass of oxygen in the sample, we'll follow these steps:

Step 1: Calculate moles of CO2CO_2 produced

The molar mass of CO2CO_2 is: Molar mass of CO2=12.01g/mol+2×16.00g/mol=44.01g/mol\text{Molar mass of CO}_2 = 12.01 \, \text{g/mol} + 2 \times 16.00 \, \text{g/mol} = 44.01 \, \text{g/mol} Now, calculate the moles of CO2CO_2 produced: Moles of CO2=4.06g44.01g/mol=0.0923mol\text{Moles of CO}_2 = \frac{4.06 \, \text{g}}{44.01 \, \text{g/mol}} = 0.0923 \, \text{mol}

Step 2: Calculate moles of carbon in the sample

In each mole of CO2CO_2, there is 1 mole of carbon. Therefore, the moles of carbon in the sample are equal to the moles of CO2CO_2 produced: Moles of carbon=0.0923mol\text{Moles of carbon} = 0.0923 \, \text{mol}

Step 3: Calculate mass of carbon in the sample

The mass of carbon in the sample can be calculated using the molar mass of carbon (12.01 g/mol): Mass of carbon=0.0923mol×12.01g/mol=1.109g\text{Mass of carbon} = 0.0923 \, \text{mol} \times 12.01 \, \text{g/mol} = 1.109 \, \text{g}

Step 4: Calculate moles of H2OH_2O produced

The molar mass of H2OH_2O is: Molar mass of H2O=2×1.008g/mol+16.00g/mol=18.016g/mol\text{Molar mass of H}_2O = 2 \times 1.008 \, \text{g/mol} + 16.00 \, \text{g/mol} = 18.016 \, \text{g/mol} Now, calculate the moles of H2OH_2O produced: Moles of H2O=1.66g18.016g/mol=0.0921mol\text{Moles of H}_2O = \frac{1.66 \, \text{g}}{18.016 \, \text{g/mol}} = 0.0921 \, \text{mol}

Step 5: Calculate moles of hydrogen in the sample

Each mole of H2OH_2O contains 2 moles of hydrogen, so the moles of hydrogen are: Moles of hydrogen=2×0.0921mol=0.1842mol\text{Moles of hydrogen} = 2 \times 0.0921 \, \text{mol} = 0.1842 \, \text{mol}

Step 6: Calculate mass of hydrogen in the sample

The mass of hydrogen in the sample can be calculated using the molar mass of hydrogen (1.008 g/mol): Mass of hydrogen=0.1842mol×1.008g/mol=0.1856g\text{Mass of hydrogen} = 0.1842 \, \text{mol} \times 1.008 \, \text{g/mol} = 0.1856 \, \text{g}

Step 7: Calculate the mass of oxygen in the sample

Now, the total mass of the sample is 2.77 g. The mass of oxygen in the sample can be calculated by subtracting the mass of carbon and hydrogen from the total mass: Mass of oxygen=2.77g(1.109g+0.1856g)=2.77g1.2946g=1.4754g\text{Mass of oxygen} = 2.77 \, \text{g} - (1.109 \, \text{g} + 0.1856 \, \text{g}) = 2.77 \, \text{g} - 1.2946 \, \text{g} = 1.4754 \, \text{g}

Final Answer:

The mass of oxygen in the sample is approximately 1.48 g.

Would you like more details or have any questions?


Here are 5 related questions:

  1. How do you determine the empirical formula of a compound from combustion data?
  2. What are the steps to find the molecular formula once you have the empirical formula?
  3. How do we use excess oxygen in combustion reactions to simplify calculations?
  4. How does the law of conservation of mass apply to combustion reactions?
  5. Why do we subtract the masses of carbon and hydrogen from the total mass to find oxygen?

Tip: Always verify molar masses of compounds and elements from the periodic table for accurate results.

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Math Problem Analysis

Mathematical Concepts

Stoichiometry
Combustion Analysis
Molar Mass

Formulas

Moles of CO2 = mass of CO2 / molar mass of CO2
Moles of H2O = mass of H2O / molar mass of H2O
Mass of oxygen = Total mass - (mass of carbon + mass of hydrogen)

Theorems

Law of Conservation of Mass

Suitable Grade Level

Grades 10-12 (High School Chemistry)