35g Mg x 1mol / 24g = 840 mol
Let empirical formula for hydrocarbon is CxHy
it will undergo combustion as
CxHy + (x + y/4) O2 ---> xCO2 + (y/2 )H2O
Given that mass of CO2 produced = 9.69 g
So moles of CO2 produced = 9.69 / 44 = 0.22 moles
So moles of carbon present = 0.22 moles
mass of H2O produced = 4.96 g
Moles of H2O produced = mass / molar mass = 4.96 / 18 = 0.28 moles
So moles of H present = 2 X 0.28 = 0.56 moles
Let us divided the moles of each with lowest value of moles
Moles of Carbon = 0.22 / 0.22 = 1 moles
moles of H = 0.56 / 0.22 = 2.55
Multiplying with two to get whole number
the ratio of carbon and hydrogen will be : C:H = 2:5
empirical formula : C2H5
Answer:
3–methyl–2–butanol
Explanation:
To name the compound, we must:
1. Identify the functional group.
2. Give the functional group of the compound the lowest possible count.
3. Locate the longest continuous carbon chain. This gives the parent name of the compound.
4. Identify the substituent group attached.
5. Give the substituent group the lowest possible count.
6. Combine the above to get the name of the compound.
Now, let us obtain the name of the compound.
1. The functional group of the compound is Alcohol i.e —OH.
2. The functional group is located at carbon 2.
3. The longest continuous carbon chain is carbon 4 i.e butane. But the presence of the functional group i.e OH will replace the –e in butane with –ol. Therefore, the compound is butanol.
4. The substituent group attached is methyl i.e CH3.
5. The substituent group is located at carbon 3.
6. Therefore, the name of the compound is:
3–methyl–2–butanol.
