A compound contains $43.58 %$ . Phosphorus & remaining oxygen. Find empirical formula: $(A M = P = 31, O = 16)$

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Solution

Step 1, Divide all the given percentage with the atomic mass.
Phosphorus $\frac{43.58}{31} = 1.4$ $1$
Oxygen $\frac{56.42}{16} = 3.5$ $2.5$
Empirical formula is $P O_{2.5}$
But empirical formula is the simplest whole number ratio.
$∴$ Empirical formula of the compound willl be $P_{2} O_{5}$ .

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Phosphorus	$\frac{43.58}{31} = 1.4$	$1$
Oxygen	$\frac{56.42}{16} = 3.5$	$2.5$

A compound contains 43.58%. Phosphorus & remaining oxygen. Find empirical formula: (AM=P=31,O=16)

Step 1, Divide all the given percentage with the atomic mass. Phosphorus 43.5831=1.41 Oxygen 56.4216=3.5 2.5Empirical formula is PO2.5But empirical formula is the simplest whole number ratio.∴ Empirical formula of the compound willl be P2O5.

A compound contains $43.58 %$ . Phosphorus & remaining oxygen. Find empirical formula: $(A M = P = 31, O = 16)$

Step 1, Divide all the given percentage with the atomic mass.
Phosphorus $\frac{43.58}{31} = 1.4$ $1$
Oxygen $\frac{56.42}{16} = 3.5$ $2.5$
Empirical formula is $P O_{2.5}$
But empirical formula is the simplest whole number ratio.
$∴$ Empirical formula of the compound willl be $P_{2} O_{5}$ .