A salt is analyzed and found to contain $0.92 \cdot g$ $0.92g$ of sodium, $1.42 \cdot g$ $1.42g$ of chlorine, and $2.56 \cdot g$ $2.56*g$ of oxygen. What is the empirical formula of this salt?

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A salt is analyzed and found to contain $0.92 \cdot g$ $0.92g$ of sodium, $1.42 \cdot g$ $1.42g$ of chlorine, and $2.56 \cdot g$ $2.56*g$ of oxygen. What is the empirical formula of this salt?

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Answer 1 · 2017-04-30T10:41:38

Well, we find the molar quantities of each constituent.........to get an empirical formula of $N a C l O_{4}$ $NaClO_4$ .

Explanation:

$Moles of sodium$ $"Moles of sodium"$ $=$ $=$ $\frac{0.92 \cdot g}{22.99 \cdot g \cdot m o l^{- 1}} = 0.0400 \cdot m o l$ $(0.92*g)/(22.99*g*mol^-1)=0.0400*mol$

$Moles of chlorine$ $"Moles of chlorine"$ $=$ $=$ $\frac{1.42 \cdot g}{35.45 \cdot g \cdot m o l^{- 1}} = 0.0400 \cdot m o l$ $(1.42*g)/(35.45*g*mol^-1)=0.0400*mol$ .

$Moles of oxygen$ $"Moles of oxygen"$ $=$ $=$ $\frac{2.56 \cdot g}{15.99 \cdot g \cdot m o l^{- 1}} = 0.160 \cdot m o l$ $(2.56*g)/(15.99*g*mol^-1)=0.160*mol$ .

Note that I DIVIDED THRU BY the ATOMIC MASS of each constituent. And now, we divide thru by the SMALLEST molar quantity, that of sodium, to get an empirical formula of $N a C l O_{4}$ $NaClO_4$ ; and this is $sodium perchlorate$ $"sodium perchlorate"$ ,

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A salt is analyzed and found to contain $0.92 \cdot g$ $0.92g$ of sodium, $1.42 \cdot g$ $1.42g$ of chlorine, and $2.56 \cdot g$ $2.56*g$ of oxygen. What is the empirical formula of this salt?

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Explanation:

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Science

Math

Humanities

... and beyond

A salt is analyzed and found to contain 0.92⋅g0.92*g of sodium, 1.42⋅g1.42*g of chlorine, and 2.56⋅g2.56*g of oxygen. What is the empirical formula of this salt?

1 Answer

Explanation:

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A salt is analyzed and found to contain $0.92 \cdot g$ $0.92g$ of sodium, $1.42 \cdot g$ $1.42g$ of chlorine, and $2.56 \cdot g$ $2.56*g$ of oxygen. What is the empirical formula of this salt?