Exercises

Exercise 1. How many moles of phosphorous are in 2.78 x 10²⁴ atoms of phosphorous?

6.02 x 10²³ atoms = 1 mole

\(\displaystyle 2.78\times\;{10^{24}\;P\;atoms}\times\frac{1\;mol\;P}{6.02\times\;10^{23}\;P\;atoms}=\;\mathbf{4.62\;moles\;P}\)

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Exercise 2. How many grams of chlorine in 32.5 moles of chlorine?

1 mole chlorine = 35.453 g

\(\displaystyle 32.5\;{moles\;Cl}\times\frac{35.453\;g\;Cl}{1\;mole\;Cl}=\;\mathbf{1.15\times\;10^{3}\;g\;Cl}\)

Exercise 3. How many selenium atoms are in 235 grams of selenium?

Roadmap: grams → moles → atoms
Two equivalences are needed.
1) 78.96 g Se = 1 mole Se and 2) 1 mole Se = 6.02 x 10²³ Se atoms

\(\displaystyle 235\;{g\;Se}\times\frac{1\;mol\;Se}{78.96\;g\;Se}\times\frac{6.02\times\;10^{23}\;Se\;atoms}{1\;mol\;Se}=\;\mathbf{1.79\times\;10^{24}\;Se\;atoms}\)

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Exercise 4. Calculate the molar mass of (NH₄)₃PO₄.

Look up the molar masses of nitrogen, hydrogen, phosphorous, and oxygen using the periodic table. Next, multiply each molar mass by the corresponding subscript in the chemical formula and then add the values.

3 x 14.0067 g/mol + 12 x 1.00794 g/mol + 30.974 g/mol + 4 x 15.9994 g/mol = 149.087 g/mol

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Exercise 5. What is the molecular mass of table sugar (also called sucrose), C₁₂H₂₂O₁₁?

Look up the molar masses of carbon, hydrogen and oxygen on the periodic table. Multiply the molar masses by the corresponding subscript in the chemical formula and add the values.

12 x 12.011 g/mol + 22 x 1.00794 g/mol + 11 x 15.9994 g/mol = 342.300 g/mol

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Exercise 6. How many moles of calcium ions are in 625 g of Ca₃(PO₄)₂? How many phosphate ions are in 625 g of Ca₃(PO₄)₂?

For the first question our roadmap is:

g Ca₃(PO₄)₂ → moles Ca₃(PO₄)₂ → moles Ca²⁺
First, calculate the mass of one mole of Ca₃(PO₄)₂

3 x 40.078 g/mol + 2 x 30.9738 g/mol + 8 x 15.9994 g/mol = 310.18 g/mol

We need two eqivalences:

1) 1 mole Ca₃(PO₄)₂ = 310.18 g
2) 1 mole Ca₃(PO₄)₂ = 3 moles Ca²⁺

\(\displaystyle 625\;{g\;Ca_{3}(PO_{4})_{2}}\times\frac{1\;mol\;Ca_{3}(PO_{4})_{2}}{310.18\;g\;Ca_{3}(PO_{4})_{2}}\times\frac{3\;mol\;Ca^{2+}}{1\;mol\;Ca_{3}(PO_{4})_{2}}=\;\mathbf{6.04\;mol\;Ca^{2+}}\)

The roadmap for the second question is:

g Ca₃(PO₄)₂ → moles Ca₃(PO₄)₂ → moles PO₄^3- → number of PO₄^3- ions

We need three eqivalences–one equivalence per arrow in the roadmap:
1) 1 mole Ca₃(PO₄)₂ = 310.18 g
2) 1 mole Ca₃(PO₄)₂ = 2 moles PO₄^3-
3) 1 mole PO₄^3- = 6.02 x 10²³ PO₄^3- ions

\(\displaystyle 625\;{g\;Ca_{3}(PO_{4})_{2}}\times\frac{1\;mol\;Ca_{3}(PO_{4})_{2}}{310.18\;g\;Ca_{3}(PO_{4})_{2}}\times\frac{2\;mol\;PO_{4}^{3-}}{1\;mol\;Ca_{3}(PO_{4})_{2}}\times\frac{6.02\times\;10^{23}\;PO_{4}^{3-}}{1\;mol\;PO_{4}^{3-}}=\;\mathbf{2.43\times\;10^{24}\;PO_{4}^{3-}}\)

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Exercise 7. Calculate the number of grams of calcium in 3.45 x 10²³ formula units (FU) of Ca(NO₃)₂.

Our roadmap: FU Ca(NO₃)₂ → moles Ca(NO₃)₂ → moles Ca → g Ca

We need 3 equivalences:

1) 6.02 x 10²³ FU Ca(NO₃)₂ = 1 mol Ca(NO₃)₂
2) 1 mol Ca(NO₃)₂ = 1 mol Ca
3) 1 mol Ca = 40.078 g

\(\displaystyle 3.45\times\;{10^{23}\;FU\;Ca(NO_3)_2}\times\frac{1\;mol\;Ca(NO_3)_2}{6.02\times\;10^{23}\;FU}\times\frac{1\;mol\;Ca}{1\;mol\;Ca(NO_3)_2}\times\frac{40.078\;g\;Ca}{1\;mol;Ca}=\;\mathbf{23.0\;g\;Ca}\)

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