Question

A 0.77 mg sample of boron reacts with oxygen to form 2.48 mg of the oxide.

Answer 1

Answer to Question #62275, Chemistry / Other

A 0.77 mg sample of boron reacts with oxygen to form 2.48 mg of the oxide.

Answer:

$1\mathrm{g} = 1000\mathrm{mg}$

$(0.77\mathrm{mg}) / (1000\mathrm{g / mg}) = 7.7*10^{-4}$ grams of B.

$$n (B) = \frac {7 . 7 \times 1 0 ^ {- 4} g}{1 0 . 8 1 \frac {g}{m o l}} = 7. 1 2 \times 1 0 ^ {- 5} m o l s o f B$$

$(2.48\mathrm{mg}) / (1000\mathrm{g / mg}) = 2.48*10^{-3}$ grams of oxide.

$$\begin{array}{l} m (O) = 2. 4 8 \times 1 0 ^ {- 3} g - 0. 7 7 \times 1 0 ^ {- 3} g = 1. 7 1 \times 1 0 ^ {- 3} g o f O \\ n (O) = \frac {1 . 7 1 \times 1 0 ^ {- 3}}{1 5 . 9 9} = 1. 0 7 \times 1 0 ^ {- 4} m o l s o f O \\ \frac {B}{O} = \frac {7 . 1 2 \times 1 0 ^ {- 5}}{1 . 0 7 \times 1 0 ^ {- 4}} = \frac {1}{1 . 5} = \frac {2}{3} \\ \end{array}$$

Obtained oxide is $\mathbf{B}_2\mathbf{O}_3$.

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