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chemistry
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Dear convert-me.com forum visitors,
Our forum has been available for many years. In September 2014 we decided to switch it to read-only mode. Month after month we saw less posts with questions and answers from real people and more spam posts. We were spending more and more resources cleaning the spam until there were less them 1 legitimate message per 100 spam posts. Then we decided it's time to stop.
All the posts in the forum will be available and searchable. We understand there are a lot of useful information and we aren't going to remove anything. As for the new questions, you can always ask them on convert-me.com FaceBook page
Thank you for being with us and sorry for any inconveniences this could caused.
2 posts
• Page 1 of 1
Re: chemistry
by Guest » Mon Jan 09, 2012 6:13 am
2012.1.8
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.
%N
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!
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.
%N
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!
- Guest
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2 posts
• Page 1 of 1