A compound of C,N and H contains 9:1:3.5 as the ratio of elements in the compound. If their MFM is 108,then calculate their EF and MF
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Answer:
In an organic compound of molar mass 108 g and C, H, N atoms are present in 9:1:3.5 by weight. What is the molecular formula of a compound?
These type of questions all of us have done is class 7 Ratio and proportion chapter where they used us to give some net amount and some ratio of 3–4 students and asked us to divide them is that particular ratio. So here they have given a total mass of 108g and u need to divide them between C, H and N in the ratio of 9 : 1 : 3.5.
Add all the ratios i.e., 9+1+3.5=13.5
Divide 108 in 13.5 part. 108–13.5=8
So each part will be of 8g
Carbon C got 9 parts then a total of 9*8= 72g
Hydrogen H got 1 parts then a total of 1*8= 8g
Nitrogen N got 3.5 parts then a total of 3.5*8= 28g
mass of each carbon atom is 12g. therefore carbon got 72g in that compound. 72/12= 6 molecules of carbon are present in that compound.
mass of each hydrogen atom is 1g. therefore hydrogen got 8g in that compound. 8/1= 8 molecules of carbon are present in that compound.
mass of each nitrogen atom is 14g. therefore nitrogen got 28g in that compound. 28/14= 2 molecules of carbon are present in that compound.
Hence C6H8N2. :-)
Explanation: