Question #c9296

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Question #c9296

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Answer 1 · 2018-01-06T12:59:10

The empirical formula of the compound is $N a_{2} S O_{4}$ $Na_2SO_4$ .

Explanation:

Suppose the mass of the compound is 100g,

Mass of sodium in the compound $\Rightarrow 100 \cdot 32.5 % = 32.5 g$ $rArr 100*32.5% = 32.5 g$

Mass of sulphur in the compound $\Rightarrow 100 \cdot 22.6 % = 22.6 g$ $rArr 100*22.6% = 22.6 g$

Mass of oxygen in the compound $\Rightarrow 100 \cdot 44.9 % = 44.9 g$ $rArr 100*44.9% = 44.9 g$

No. of moles of sodium in the compound $\Rightarrow \frac{32.5}{23} = 1.4 m o l$ $rArr 32.5/23 = 1.4 mol$

No. of moles of sulphur in the compound $\Rightarrow \frac{22.6}{32} = 0.7 m o l$ $rArr 22.6/32 = 0.7 mol$

No. of moles of oxygen in the compound $\Rightarrow \frac{44.9}{16} = 2.8 m o l$ $rArr 44.9/16 = 2.8 mol$

Mole ratio $\Rightarrow \times \times N a \times \times : \times \times S \times \times x : \times \times O$ $rArr color(white)(xxxx) Nacolor(white)(xxxx) : color(white)(xxxx)S color(white)(xxxxx) : color(white)(xxxx)O$

$\times x . \times x \Rightarrow \times x 1.4 m o l \times x : \times x 0.7 m o l \times x : \times x 2.8 m o l$ $color(white)(xxx.xxx) rArr color(white)(xxx)1.4mol color(white)(xxx) : color(white)(xxx) 0.7mol color(white)(xxx) : color(white)(xxx) 2.8mol$

|Dividing through by the smallest ratio..|

$\times x . \times x \Rightarrow \times x \frac{1.4 m o l}{0.7 m o l} \times x : \times x \frac{0.7 m o l}{0.7 m o l} \times x : \times x \frac{2.8 m o l}{0.7 m o l}$ $color(white)(xxx.xxx) rArr color(white)(xxx)(1.4mol)/(0.7mol) color(white)(xxx) : color(white)(xxx) (0.7mol)/(0.7mol) color(white)(xxx) : color(white)(xxx) (2.8mol)/(0.7mol)$

$\times x . \times x \Rightarrow \times . \times x 2 \times \times x : \times \times \times 1 \times \times : \times \times x 4$ $color(white)(xxx.xxx) rArr color(white)(xx.xxx) 2 color(white)(xxxxx) : color(white)(xxxxxx) 1 color(white)(xxxx) : color(white)(xxxxx) 4$

Thus, the empirical formula of the compound is $N a_{2} S O_{4}$ $Na_2SO_4$ .

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Question #c9296

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