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  1. Problems
  2. /A sugar of molar mass 150 g/mol contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass…
MediumChemistryHigh School (11–12)Organic Compounds with Functional Groups › BiomoleculesNumerical

A sugar of molar mass 150 g/mol contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass…

StemPy LabContributed by StemPy Lab

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A sugar of molar mass 150 g/mol150\ \text{g/mol}150 g/mol contains 40.0%40.0\%40.0% carbon, 6.7%6.7\%6.7% hydrogen and 53.3%53.3\%53.3% oxygen by mass. Take C as 12.011 g/mol12.011\ \text{g/mol}12.011 g/mol, H as 1.008 g/mol1.008\ \text{g/mol}1.008 g/mol and O as 15.999 g/mol15.999\ \text{g/mol}15.999 g/mol. Find the empirical formula, then the molecular formula. If the sugar is an aldopentose with an unbranched chain, how many stereoisomers does it have?

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Need a reminder? The key ideaShowHide

A carbohydrate is a polyhydroxy aldehyde or ketone. An aldose has n−2n - 2n−2 stereocentres and a ketose has n−3n - 3n−3, where nnn is the number of carbons in the chain. The number of stereoisomers is 2centres2^{\text{centres}}2centres, and half of them belong to the D series. An empirical formula comes from the moles of each element in 100 g.

The worked solution

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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.

Plan. Moles of each element give the empirical formula. The molar mass gives the multiplier.

Step 1. Moles in 100 g. Carbon: 40.0/12.011=3.33 mol40.0 / 12.011 = 3.33\ \text{mol}40.0/12.011=3.33 mol. Hydrogen: 6.7/1.008=6.65 mol6.7 / 1.008 = 6.65\ \text{mol}6.7/1.008=6.65 mol. Oxygen: 53.3/15.999=3.33 mol53.3 / 15.999 = 3.33\ \text{mol}53.3/15.999=3.33 mol.

Step 2. Ratio. Dividing by 3.33 gives C : H : O = 1 : 2 : 1. The empirical formula is CH2O\text{CH}_{2}\text{O}CH2​O, with a molar mass of 30.03 g/mol30.03\ \text{g/mol}30.03 g/mol.

Step 3. Molecular formula. 150/30.03=5150 / 30.03 = 5150/30.03=5. The formula is C5H10O5\text{C}_{5}\text{H}_{10}\text{O}_{5}C5​H10​O5​, so there are 5 carbon atoms.

Step 4. Stereoisomers. An aldopentose has 3 stereocentres, so 23=82^3 = 823=8 stereoisomers.

Answer: C5H10O5\text{C}_{5}\text{H}_{10}\text{O}_{5}C5​H10​O5​, with 8 stereoisomers if it is an aldopentose.

Check. The hydrogen is twice the oxygen in the ratio, as in every carbohydrate of this kind: Cn(H2O)n\text{C}_\text{n}(\text{H}_{2}\text{O})_\text{n}Cn​(H2​O)n​.

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