I have always loved the way Greek prefixes make shape names & units of measurement so easy to remember, and they were my in-road to this problem. It’ll be worth your time to learn them if you don’t already know them!

See the original problem before the solution?

**Can you match the names to the formulae?**

C_{3}H_{8}, C_{4}H_{6}, C_{3}H_{4}, C_{4}H_{8}, C_{7}H_{14}, C_{2}H_{2}

Heptene, Butyne, Propane, Butene, Ethyne, Propyne

There are many ways to solve this without any chemistry knowledge, but as a Math teacher I knew heptagons are 7-sided polygons so this was my break-in point. There was only a single 7 in all the formulae so I was in luck! This told me that the **prefix **of the name is determined by the **number of Carbon atoms** in the formula.

**Hept**ene = **C**_{7}H_{14}

There are 3 different types of **endings for the chemical names** (**suffixes**) which allowed me to group them into 3 categories:

**-enes**

Heptene & butene

**-anes**

Propane

–**ynes**

Butyne, ethyne & propyne

**I tried to draw diagrams of the compounds but did not get very far as I could not remember how their bonds work!**

So, if the number of carbons in the compound determines its prefix, it must be the ratio of carbons to hydrogens that determines the suffix (otherwise there would have to be another number in the name (e.g. buhexane)).

**Analysing the ratio of carbon ** **to hydrogen atoms in the formulae **revealed some patterns:

- there were twice as many Hs as Cs in C
_{7}H_{14}& C_{4}H_{8} - C
_{3}H_{8}, C_{4}H_{6}, C_{3}H_{4}have more Hs than Cs, (but**not**twice as much) - C
_{2}H_{2}has equal numbers of H and C

There are only 2 compounds in the -enes group so I figured that if C_{7}H_{14} is **heptene**, then **C**_{4}**H**_{8}** **must be **butene**, the only other -ene This implies that compounds having **4 carbons should be named ‘bu-something’.**

I realised I could not group these two compounds together (C_{3}H_{8 & }C_{3}H_{4}), because the grouping was to determine the ratio & hence suffix, (with the 3 Cs giving the prefix). So I put C_{3}H_{8} into its own group because it had more than 2 Hydrogens for every Carbon. Then I put C_{2}H_{2} into the group because it had to go somewhere and I only wanted 3 groups. Not very scientific I know but it worked!

Here are the groups by ratio of C to H, to hopefully work out the suffixes.

**Group 1**:

C_{7}H_{14,} C_{4}H_{8}

**Group 2: **

C_{3}H_{8} **Group 3: **

C_{4}H_{6}, C_{3}H_{4,} C_{2}H_{2}

So the group with just a single compound must be the only -ane, so thereore**:C**

_{3}

**H**

_{8}

**= propane**

Group 3 must be the three “-ynes”.

The prefix ‘pro’ must mean 3 carbons, as propane had 3, so:

C_{3}H_{4} = propyne

Above we figured that **4 carbons would be named bu-something **so:

C_{4}H_{6} = butyne

There is only one left!

C_{2}H_{2} = ethyne

The rest was fairly easy. Let me know if you need any help with it. My feet are too cold to write it now – I need to go for a walk!

Thanks for reading. Sincerely.

## LINKS

- Tanya Khovanova’s fascinating Math blog
- Russian Internet Linguistics Olympiad
- how how is it inside a hot-air balloon?

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