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Question: \({{C}_{3}}{{H}_{7}}OH\) is propanol. Suppose you have an 85.9 g sample of propanol. How many carbon...

C3H7OH{{C}_{3}}{{H}_{7}}OH is propanol. Suppose you have an 85.9 g sample of propanol. How many carbon atoms are in the sample?

Explanation

Solution

As we know that the molecular formula of propanol is C3H7OH{{C}_{3}}{{H}_{7}}OH. In order to solve this question, we will first calculate its molecular mass. And we should have knowledge of the mole concept.

Complete step by step answer:
- Let us solve the question, firstly as we know that the molar masses of carbon, hydrogen, and oxygen are 12 g/mol, 1 g/mol, 16 g/mol, we will add up all these values:
M= (3×\times 12) + (8×\times 1) + (1×\times 16)
M= 60 g/mol
- As we know that elements are made up of substances that are in different proportions, this is because they have different masses. We will find out how much propanol is actually present in carbon in the terms of percentage.
- We should remember that 36 g of that is carbon
60g10060g\to 100 %
36gx36g\to x%
x=60\Rightarrow x=60%
- As of now we know that any given sample of propanol is 60% carbon. We are being provided with the information that 85.9 g is the total mass of propanol, then 60% of that must be carbon.
85.9g10085.9g\to 100 %
yg60'y'g\to 60 %
y=51.4g\Rightarrow y = 51.4g
- Now, let us use the mole concept. As molar mass is the mass of a mole of something, hence we can say 6.022×10236.022\times {{10}^{23}} carbon atoms have a total mass of 12 g. Now, we can find that how many atoms of carbon have a total mass of 51.54 g.
6.022×102312g6.022\times {{10}^{23}}\to 12g
z atoms51.54g\to 51.54g
z=2.58×1024\Rightarrow z=2.58\times {{10}^{24}}atoms
- Hence, we can conclude that there are 2.58×10242.58\times {{10}^{24}} carbon atoms in the sample.

Note: - We should not get confused in the terms mole and molecule. As mole is used to measure the quantity of a substance and is having the unit mol whereas, the molecule is used to name a combination of some atoms and can be measured as the number of ‘mol’ present.