What is the empirical formula of a phosphoric acid that contains 0.3086 g of hydrogen, 3.161 g of phosphorus, and 6.531 g of oxygen?

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What is the empirical formula of a phosphoric acid that contains 0.3086 g of hydrogen, 3.161 g of phosphorus, and 6.531 g of oxygen?

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Answer 1 · 2016-04-29T22:31:20

$H_{2}$ $H_2$ $P_{4}$ $P_4$ $O_{1}$ $O_1$

Explanation:

In order to calculate the Empirical formula , we will assume that we have started with 10 g of the compound. On decomposition , we obtained 0.3086 g of Hydrogen , 3.161 g of Phosphorus, and 6.531 g of Oxygen.

*Step 1* Calculating the number of moles of different elements

$n$ $n_$ $H$ $H$ = mass of H/ molar mass of H

$n$ $n_$ $H$ $H$ = 0.3086 g / 1 g $m o l^{- 1}$ $mol^(-1)$

$n$ $n_$ $H$ $H$ = 0.3086 mol

$n$ $n_$ P = mass of P/ molar mass of P

$n$ $n_$ P $= 3.161 \frac{g}{31} g$ $= 3.161 g / 31 g$ mol^(-1)#

$n$ $n_$ $P$ $P$ = 1.02 mole

$n$ $n_$ $O$ $O$ = mass of O / molar mass of O

$n$ $n_$ $O$ $O$ = 6.531 g / 32 g $m o l^{- 1}$ $mol^(-1)$

$n$ $n_$ $O$ $O$ = 0.2 mol

Calculating the Simple Ratio of the moles

H= 0.3806/ 0.2 =4 , P = 1.02 / 0.2 =5 , O= 0.2/0.2 =1

$H_{2}$ $H_2$ $P_{4}$ $P_4$ $O_{1}$ $O_1$

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What is the empirical formula of a phosphoric acid that contains 0.3086 g of hydrogen, 3.161 g of phosphorus, and 6.531 g of oxygen?

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