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Topic: empirical formluaes  (Read 4469 times)

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Sobana

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empirical formluaes
« on: October 16, 2005, 11:12:10 AM »
hiya can anyone help me about how I would go about answering this question?

A farmer buys a bag of fertiliser which is a mixture of ammonium sulphate ((NH4)2SO4) and potassium nitrate (KNO3). the bag weighs 20kg and 30% of this mass is ammonium sulphate. calculate the total mass of nitrogen in the bag of fertilizer.

if we say the mass for KNO3 is 70%

i know that the reletive atomic mass for:

k= 39.1
n=14
ox3= 48

how would i get the individual mass for the elements?

anything will be much appreciated

Offline mike

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Re:empirical formluaes
« Reply #1 on: October 16, 2005, 06:35:05 PM »
Work out the number of moles of each compound, then multiply by the molecular weight of the element.

30% of 20kg = 6000g

M (NH4)2SO4 = 132.13 g.mol-1

n = m/M = 6000/132.13 = 45.4 moles
---------------------------------------------
M (N) = 14.01 g.mol-1

m (N) = n.M = 45.4 x 14.01 = 636.2 g (of N in the bag due to ammonium sulfate)

work out the rest of the N mass then add together..
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Offline Donaldson Tan

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Re:empirical formluaes
« Reply #2 on: October 19, 2005, 08:06:22 AM »
what is the % mass composition of nitrogen in KNO3?

molar mass of KNO3 = 39.1 + 14 + 3X16 = 101.1
% mass composition of nitrogen in KNO3 = 14/101.1 X 100% = 13.8%

do the same for ammonium sulphate.

u know the mass of ammonium sulphate and potassium nitrate in the fertiliser, so you can work out the total mass of Nitrogen in the fertiliser.
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