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Postby bigballer » Sun Sep 25, 2011 8:12 am

What is the % nitrogen by weight in 2.114 mol of ammonium sulfide? Answer in units of %.

Re: chemistry

Postby Guest » Mon Jan 09, 2012 6:13 am

ammonium nitrate (NH4)(NO3), molar mass 80.052 g/mol
2.114 mol x 80.052 g/mol = 169.2299 g

The molar mass of elements is found by looking at the atomic mass of the element on the periodic table
Nitrogen = 14.0067 atomic wt
Hydrogen = 1.00794
Oxygen = 15.9994

so two N + four H + three O = 80.04336; Nitrogen makes up two atoms in the compound ammonium nitrate so
2 x 14.0067 = 28.0134 grams per mole, there are 2.114 moles so there is 59.2203 g on N in that sample. (2.114 moles x 28.0134 g/mol = 59.2203 g of N)

2.114 mole x 80.04336 g/mol (my calculated molar mass 80.04336 vs the reference value 80.052) = 169.211 g in this sample.
59.2203 / 169.211 = 0.3499778 so the %N is about 34.998 % of the sample
(Oxygen weighing more and there being more atoms than Nitrogen, Oxygen makes up most of the % in that compound.)

A possible concern is Nitrogen is diatomic meaning that each molecule of the element has two atoms of that element stuck together. So hopefully one does not have to half the atomic weight value of Nitrogen from 14.0067 to 7.0033. That speculation would rather complicate this calculation!

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