break up the carbon dioxide to C and O
2.93g>>(12/44)*2.95=.80gC;
break up the water into H and O
1.80g>>1.8*(2/18)=.2gH;
now total them up, change to moles
C=,8g=.8/12=.066moles
H=.2g=.2/1= .2 moles
Notice that the sum of H and C=1.0g, so all O was atmospheric
now divide by the least of the above (.066)
C=1
H=.2/.066=3
empirical: CH3
a hydrocarbon is burnt completely in excess oxygen. it is found that 1.00g of the hydrocarbon gives 2.93g carbon dioxide and 1.80g water. find the empirical formula of the hydrocarbon (H=1,C=12,O=16)
1 answer
1 answer