Back
to top
as chemistry
Watch this threadPage 1 of 1
Go to first unread
Skip to page:
uzer
Badges:
9
Rep:
?
You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#1
when K is burnt in excess O2, a compound is produced that contains 54.9% K.
Calculate the percentage of oxygen present & hence calculate the empirical formula of this compound
Calculate the percentage of oxygen present & hence calculate the empirical formula of this compound
0
reply
Nirgilis
Badges:
14
Rep:
?
You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#2
Dominic1905
Badges:
0
Rep:
?
You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#3
Report
#3
(Original post by uzer)
when K is burnt in excess O2, a compound is produced that contains 54.9% K.
Calculate the percentage of oxygen present & hence calculate the empirical formula of this compound
when K is burnt in excess O2, a compound is produced that contains 54.9% K.
Calculate the percentage of oxygen present & hence calculate the empirical formula of this compound
The key information in this question is that the K is being burnt in an excess of O2; this is important because it means that O2 isn't a limiting reagent, there's more than enough of it around to react fully with the K, so we can assume that the rest of the compound is oxygen.
Therfore % Oxygen = 100% - K%
100%-54.9%=45.1%
Now we know the percentage composition of each component of the compound we can divide the relevant percentage by the molar mass of that element and work out the empirical formula:
K
54.9%/39=1.4
O
45.1%/16=2.8
Dividing by the lowest number give the ratio in its simplest form:
1.4/1.4=1 therefore we have 1 K atom present
2.8/1.4=2 therefore we have 2 O atoms present
This means the empirical formula is KO2
0
reply
X
Page 1 of 1
Go to first unread
Skip to page:
Quick Reply
