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Butane, C₄H₁₀, has a molar mass of 58 g/mol…
Contributed by StemPy LabTry it on paper first. A hint, the key idea and the worked solution are here when you need them.
Butane, , has a molar mass of 58 g/mol. Alkanes form a homologous series. Find the molar mass of the next alkane, and of the alkane with eight carbon atoms. Take C = 12 g/mol and H = 1 g/mol.
Need a reminder? The key ideaShow
Alkanes , alkenes , alkynes . A homologous series differs by , 14 u. Name = prefixes (branches, in alphabetical order) + parent chain + suffix (principal group). Use the longest chain, with the principal group and the multiple bonds in it. Number to give the lowest locants at the first point of difference: principal group first, then multiple bonds, then branches. Seniority: acid, ester, amide, nitrile, aldehyde, ketone, alcohol, amine, alkene, alkyne.
The worked solution
Have you tried it?
Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Step 1. One step. A group has a molar mass of g/mol.
Step 2. The next alkane. Pentane has g/mol.
Step 3. Eight carbons. Four steps up from butane adds g/mol. So g/mol. Check with the formula : g/mol.
Answer: 72 g/mol and 114 g/mol.
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