Question

Oblicz skład procentowy masowy poniższych związków chemicznych
a) metan CH4
b) amoniak NH3
c) tlenek fosforu (V) P4O10
d) metanol CH3OH
e) kwas mrówkowy HCOOH
f) ,, nitrogliceryna " C3H5 (ONO2)3

Answer 1

A)mCH4=12u+4*1u=16u
12u-x%
16u-100%
x=75%C
4u-x%
16u-100%
x=25%H
b)mNH3=14u+3*1u=17u
14u-x%
17u-100%
x=82,35%N
3u-x%
17u-100%
x=17,65%H
c)mP4O10=4*31u+10*16u=284u
124u-x%
284u-100%
x=43,66%P
160u-x%
284u-100%
x=56,34%O
d)mCH3OH=12u+4*1u+16u=32u
12u-x%
32u-100%
x=37,5%C
4u-x%
32u-100%
x=12,5%H
16u-x%
32u-100%
x=50%O
e)mHCOOH=2*1u+12u+2*16u=46u
2u-x%
46u-100%
x=4,35%H
12u-x%
46u-100%
x=26,09%C
32u-x%
46u-100%
x=69,56%O
f)mC3H5(ONO2)3=3*12u+5*1u+9*16u+3*14u=227u
36u-x%
227u-100%
x=15,86%C
5u-x%
227u-100%
x=2,2%H
144u-x%
227u-100%
x=63,44%O
42u-x%
227u-100%
x=18,5%N

Answer 2

CH4

mcz - 12 + 4 = 16 u

16 --- 100%
12 ------X

X = 75 % - C

%O = 100% - 75 % = 25 %

NH3

Mcz = 14 + 3 = 17 u

17 ---- 100%
14 -----X

X = 82,3 % - N

% O = 100% - 82,3 % = 17,7 %

P4O10

mcz= 4 x 31 + 10 x 16 = 284 u

284 ---- 100%
124 -------- X

X = 43,7 % - P

%O = 100% = 43,7 % = 56,3 %

HCOOH

mcz = 2 + 12 + 2 x 16 = 46 u
46 --- 100%
12 -----X

x = 26,1 % C

46 --- 100%
32------ X

X = 69,6 % - O

% H = 100% - ( 69,6 + 26,1 ) = 4,3 %

C3H5(ONO2)3

mcz= 3 x 12 + 5 + 9 x 16 +3 x 14 = 36 + 5 +144 + 42 = 227 u

227 ---- 100%
36 ------ X

X = 15,8 %  - C

227 --- 100%
5 -------- X

X = 2,2 % - H

227 - 100%
9 x 16 ---X

X = 63,4 % - O

% N = 100% - ( 63,4 + 2,2 + 15,8 ) = 100% - 81,4 % = 18,6 %