Glucose, a simple sugar, is the main component of intravenous feeding liquids. Its common name is dextrose. It is also one of the products of carbohydrate metabolism. An elemental analysis showed that glucose contains only `C, H`, and `O`. `A 0.1014 g` sample of purified glucose was burned in a `C - H` combustion system to produce `0.1486 g` of `CO_(2)` and `0.0609 g` of `H_(2)O`. Determine the masses of `C, H`, and `O` in the sample and the percentages of these elements in glucose.
Strategy:
Step `1`. and `2`. First calculate the masses of `C` and `H` as illustrated in Example `13.1`.
Step `3`. The rest of the sample must be `O` as glucose has been shown to contain only `C, H`, and `O`. Therefore, subtract the masses of `C` and `H` from the total mass of sample.
Step `4`. Calculate the percentage by mass for each element.

Explanation

Step 1. ? `gC = 0.1486 g CO_(2) xx (12.01 g C)/(44.01 g CO_(2))`
`= 0.04055 g`
Step 2. ? `gH = 0.0609 g H_(2)O xx (2.016 g H)/(18.02 g H_(2)O)`
`= 0.0681 g H`
Step 3. ? `g O = (0.1014 g "sample") - (0.04055 g C + 0.00681 g H)`
`= 0.0540 g O`
Step 4 `% C = (0.04055 g C)/(0.1014 g "sample") xx 100% = 39.99% C`
`% H = (0.00681 g H)/(0.1014 g "sample") xx 100% = 6.72% H`
`% O = (0.0540 g O)/(0.1014 g "sample") xx 100% = 53.2% O`
Total `= 99.9%`
The mass of `O` in the sample is calculated by difference.