# Answers to Percent Composition and Empirical Formula

1.

1. 1 mole KClO3 = (39.1 g K) + (35.4 g Cl) + (3 x 16.0 = 48.0 g O) = 122.5 g

$$\mathrm{\%\: K = \dfrac{39.1}{122.5} = 31.9\%\: K}$$;

$$\mathrm{\%\: Cl = \dfrac{35.4}{122.5} = 28.9 \%\: Cl}$$;

$$\mathrm{\%\: O = \dfrac{48.0}{122.5} = 39.2\%\: O}$$

1. calcium hydroxide = Ca(OH)2

1 mole Ca(OH)2 = (40.08 g Ca) + (2 x 16.0 = 32.0 g O) + (2 x 1.008 = 2.02 g H) = 74.1 g

$$\mathrm{\%\: Ca = \dfrac{40.08}{74.1} = 54.1\%\: Ca}$$;

$$\mathrm{\%\: O = \dfrac{32.0}{74.1} = 43.2\%\: O}$$;

$$\mathrm{\%\: H = \dfrac{2.02}{74.1} = 2.73\%\: H}$$

2.

1. In 100 g of the compound:

$$\mathrm{41.3\: g\: C \times \left(\dfrac{1\: mol\: C}{12.0\: g\: C}\right)= 3.44\: mol\: C}$$

$$\mathrm{10.4\: g\: H \times \left(\dfrac{1\: mol\: H}{1.01\: g\: H}\right) =10.4\: mol\: H}$$

$$\mathrm{48.2\: g\: N \times \left(\dfrac{1\: mol\: N}{14.0\: g\: N}\right)=3.44\: mol\: N}$$

Ratio: $$\mathrm{\textrm{C:}\:\: \dfrac{3.44}{3.44} = 1}$$; $$\mathrm{\textrm{H:}\:\: \dfrac{10.4}{3.44} = 3}$$; $$\mathrm{\textrm{N:}\:\: \dfrac{3.44}{3.44} = 1}$$;              CH3N

1. In 100 g of the compound:

$$\mathrm{1.92\: g \times\left(\dfrac{1\: mol\: Mn}{54.9\: g}\right) = 0.350\: mol\: Mn}$$

$$\mathrm{1.12\: g \times\left(\dfrac{1\: mol\: O}{16.0\: g}\right)=0.700\: mol\: O}$$

Ratio: $$\mathrm{\textrm{Mn:}\:\: \dfrac{0.0350}{0.0350} = 1}$$; $$\mathrm{\textrm{O:}\:\: \dfrac{0.0700}{0.0350} = 2}$$;                       MnO2

1. In 100 g of the compound:

$$\mathrm{6.51\: g\: Cu \times\left(\dfrac{1\: mol\: Cu}{63.5\: g\: C}\right) = 1.03\: mol\: Cu}$$

$$\mathrm{32.8\: g\: O \times\left(\dfrac{1\: mol\: O}{16.0\: g\: O}\right) = 2.05\: mol\: O}$$

$$\mathrm{2.1\: g\: H \times\left(\dfrac{1\: mol\: H}{1.0\: g\: H}\right) = 2.1\: mol\: H}$$

Ratio: $$\mathrm{\textrm{Cu:}\:\: \dfrac{1.03}{1.03} = 1}$$; $$\mathrm{\textrm{O:}\:\: \dfrac{2.05}{1.03} = 2}$$; $$\mathrm{\textrm{N:}\:\: \dfrac{2.1}{1.03} = 2}$$;

Cu(OH)2 (it contains copper, so it's ionic)  name = copper(II) hydroxide

1. In 100 g of the compound:

$$\mathrm{1.56\: g\: C \times \left(\dfrac{1\: mol\: C}{12.0\: g\: C}\right) = 0.130\: mol\: C}$$

$$\mathrm{0.333\: g\: H \times \left(\dfrac{1\: mol\: H}{1.01\: g\: H}\right) = 0.333\: mol\: H}$$

$$\mathrm{2.08\: g\: O \times \left(\dfrac{1\: mol\: O}{16.0\: g\: O}\right) = 0.130\: mol\: O}$$

$$\mathrm{0.910\: g\: N \times \left(\dfrac{1\: mol\: N}{14.0\: g\: N}\right) = 0.0650\: mol\: N}$$

Ratio: $$\mathrm{\textrm{C:}\:\: \dfrac{0.130}{0.0650} = 2}$$; $$\mathrm{\textrm{H:}\:\: \dfrac{0.333}{0.0650} = 5}$$; $$\mathrm{\textrm{O:}\:\: \dfrac{0.130}{0.0650} = 2}$$;

$$\mathrm{\textrm{N:}\:\: \dfrac{0.0650}{0.0650} = 1}$$;                        C2H5O2N