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Difficulty: Very hardThe Mole Concept, Avogadro's Constant, and Molar Mass

A sample of pure ammonium trioxocarbonate(IV), (NH4)2CO3(\text{NH}_4)_2\text{CO}_3, is determined to contain 3.6×10243.6 \times 10^{24} hydrogen atoms. What is the total mass of oxygen present in this sample?

[Relative atomic masses: H=1\text{H} = 1, C=12\text{C} = 12, N=14\text{N} = 14, O=16\text{O} = 16; Avogadro's constant NA=6.0×1023 mol−1N_A = 6.0 \times 10^{23} \text{ mol}^{-1}]

  1. A
    12 g12\text{ g}
  2. 36 g36\text{ g}Answer
  3. C
    72 g72\text{ g}
  4. D
    2.25 g2.25\text{ g}

Answer

36 g36\text{ g}
The correct answer is 36 g36\text{ g}. Converting 3.6×10243.6 \times 10^{24} hydrogen atoms using Avogadro's constant gives 6.0 moles6.0\text{ moles} of hydrogen atoms. Because each formula unit of (NH4)2CO3(\text{NH}_4)_2\text{CO}_3 contains 88 hydrogen atoms and 33 oxygen atoms, the mole ratio of O to H is 3:83:8, giving 2.25 moles2.25\text{ moles} of oxygen atoms. Multiplying 2.25 moles2.25\text{ moles} by the molar mass of oxygen (16 g/mol16\text{ g/mol}) yields 36 g36\text{ g}.

Step-by-Step Solution

1
Calculate the total number of moles of hydrogen atoms from the given particle count.
Moles of H atoms=3.6×10246.0×1023 mol−1=6.0 moles\text{Moles of H atoms} = \frac{3.6 \times 10^{24}}{6.0 \times 10^{23} \text{ mol}^{-1}} = 6.0\text{ moles}.
Dividing particle count by Avogadro's constant gives the amount in moles.
2
Determine the number of hydrogen atoms and oxygen atoms in one formula unit of (NH4)2CO3(\text{NH}_4)_2\text{CO}_3.
One formula unit contains 2×4=82 \times 4 = 8 hydrogen atoms and 33 oxygen atoms.
The subscript 2 outside the ammonium group (NH4)(\text{NH}_4) multiplies both N and H inside.
3
Calculate the moles of oxygen atoms present in the sample.
Moles of O atoms=6.0 moles of H×3 mol O8 mol H=2.25 moles of O\text{Moles of O atoms} = 6.0\text{ moles of H} \times \frac{3\text{ mol O}}{8\text{ mol H}} = 2.25\text{ moles of O}.
The mole ratio of O to H in the chemical formula is 3:83 : 8.
4
Convert the moles of oxygen atoms to mass.
Mass of O=2.25 mol×16 g/mol=36 g\text{Mass of O} = 2.25\text{ mol} \times 16\text{ g/mol} = 36\text{ g}.
Mass is obtained by multiplying the number of moles by the molar mass of oxygen.

Key Concept

Stoichiometric mole relationships between constituent elements in a chemical compound using Avogadro's constant and molar mass.
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Topics

The Mole Concept, Avogadro's Constant, and Molar Mass

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