what is the empirical formula of an oxide of manganese contains 69.6%o of manganese?( mn=55,o=16
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Answer:
Since
%
litterally means "per hundred" let's take a 100g sample.
Explanation:
We convert the mass ratio into mol ratio:
Then 72g of this is Manganese.
The molar mass of
M
n
=
55
, so
72
g
↔
72
55
=
1.31
m
o
l
of
M
n
The 28g of Oxygen (molar mass
O
=
16
) breaks down to:
28
g
↔
28
16
=
1.75
m
o
l
of
O
The mol ratio
M
n
÷
O
=
1.31
÷
1.75
=
0.75
÷
1
=
3
÷
4
Empirical formula:
M
n
3
O
4
Note:
This works if you take any other sample size, but then there are more calculations to do.
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