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OpenStudy (anonymous):
Determine the % composition of ammonium nitrate: NH4NO3
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OpenStudy (anonymous):
precent composition i think. What i typed is the question(this time no typos)
OpenStudy (chmvijay):
can you tell molecular weigh of NH4NO3
OpenStudy (anonymous):
80.052
OpenStudy (chmvijay):
how many diffrernt elements are present in NH4NO3 can you tell
OpenStudy (anonymous):
4
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OpenStudy (chmvijay):
what are those ??
OpenStudy (anonymous):
N - nitrogen
H4 - hydrogen
N - nitrogen
O3 - oxygen
OpenStudy (chmvijay):
can you tell the atomic mass of N, O and H
OpenStudy (anonymous):
14.007 - N
15.999 - O
1.008 - H
OpenStudy (chmvijay):
now in 80 gram of NH4NO3 you have 2*14 gramof N right
then for 100 gram =2*14*100 /80=
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OpenStudy (chmvijay):
can you calculate this similarly do for the O and H too
OpenStudy (anonymous):
2 x 14 x 100 / 80 = 35
2 x 16 x 100 / 80 = 40
2 x 1 x 100 / 80 = 2.5
OpenStudy (chmvijay):
you should not take 2 for all that is 3 for O and 4 for hydrogen :)
OpenStudy (anonymous):
oh so
2 x 14 x 100 / 80 = 35
3 x 16 x 100 / 80 = 60
4 x 1 x 100 / 80 = 5
OpenStudy (chmvijay):
yaaa now total is 100 % right
the compostion of NH4 NO# is 35 % N ,60% O and 5 % H
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