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# Empirical & Molecular Formula

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When phosphorus is burnt in a free supply of air, diphosphorous pentoxide is formed. During later stages of combustion in a limited supply of air, the flame becomes larger greenish and flickering. Another phosphorus oxide is formed which is a white waxy crystalline solid which melts @ 23.8 degrees C and becomes a gas @ 176 degrees C.

a) When the composition of the waxy crystalline solid is determined it is found to contain 56.4% phosphorus by mass, the remainder being oxygen. Use the % composition together with the atomic mass relative to oxygen to obtain the empirical formula of the compound.

At 200 degrees C and a pressure of 760 mmHg, a sample of the oxide with a mass of 0.568g has a volume of 99.5cm3

b) Calculate the volume, in cm3 that the sample would have @ STP, assuming that it remains gaseous.

c) Calculate the density in g/litre, of the gaseous oxide of phorphorus @ STP.

d) Determine the molecular formula of the oxide of phosphorous when it is in the gaseous state.

https://brainmass.com/chemistry/stoichiometry/empirical-molecular-formula-17839

#### Solution Preview

a) Let's just assume that we have 100 g of crystalline solid, that is, 56.4 g phosphorous (P) and 43.6 g oxygen (O). This assumption can be made as we just want to work with percentages and ratios, and it makes things much easier. By using the atomic weights, we can calculate the number of moles of P and O:

P: 56.4 g/30.974 g/mol = 1.82 moles
O: 43.6 g/15.999 g/mol = 2.725 moles

This means a P1.82O2.725 formula. To get the empirical formula, we need to get whole numbers, so lets divide both numbers by the smaller number 1.82. This gives: P1O1.5, which is equivalent to P2O3.
This makes sense, as phosphorous is in the +5 oxidation state in P2O5, but it is in a lower oxidation state in P2O3 (+3), which is a common ...

#### Solution Summary

The problem is solved using the empirical and molecular formula with details provided.

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