Empirical and Molecular Forumla calculation

[Oblivion77]

Homework Statement

Reductic acid contains 52.63% carbon, 5.30% hydrogen and 42.07% oxygen by mass.
a) Determine the empirical formula of reductic acid

b) If 1.23 g of reductic acid contains 3.25x10^22 atoms of carbon, what is the molecular formula of reductic acid ?

c) Calculate the mass of one molecule of reductic acid.

Homework Equations

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The Attempt at a Solution

I already found that the empirical formula and found that to be C5H6O3. But I can't seem to get the other parts. Thanks for the help.

 

[GCT]

Use the molecular mass of the empirical formula to find the number of carbon atoms that there would be in 1.23 g and then compare this to the actual number of carbons given in the problem

e.g.

1.23 g ( moles / g = MW of C5H6O3 ) = moles of C5H6O3

convert this to atoms of C5H6O3 and then this to atoms of Carbon = x

3.25 X 10 ^ 22 atoms of carbon / x = the number factor to multiply your empirical formula to find the molecular formula.




For the next part simply use Avogadro's number.

 

[Blueshift5]

First, let's review the definitions of empirical and molecular formulas. An empirical formula is the simplest whole number ratio of atoms in a compound, while a molecular formula gives the exact number of each type of atom in a molecule. To determine the molecular formula, we need to know the molar mass of the compound.

a) To find the empirical formula, we can assume a 100 g sample of reductic acid, which would give us 52.63 g of carbon, 5.30 g of hydrogen, and 42.07 g of oxygen. We then need to convert these masses to moles by dividing by the molar masses of each element:

52.63 g C x (1 mol C/12.01 g C) = 4.38 mol C
5.30 g H x (1 mol H/1.01 g H) = 5.25 mol H
42.07 g O x (1 mol O/16.00 g O) = 2.63 mol O

We can then find the simplest whole number ratio by dividing each of these values by the smallest number of moles, which in this case is 2.63 mol O. This gives us an empirical formula of C2H2O.

b) To find the molecular formula, we need to know the molar mass of reductic acid. To do this, we can use the molar mass of the empirical formula (C2H2O) and compare it to the molar mass of the actual compound, which is given by the problem.

Molar mass of C2H2O = (2 x 12.01 g/mol) + (2 x 1.01 g/mol) + (1 x 16.00 g/mol) = 42.04 g/mol

We can then calculate the molar mass of the actual compound using the given percentages:

(52.63 g C/0.5263) + (5.30 g H/0.0530) + (42.07 g O/0.4207) = 100 g/mol

This means that the actual molar mass is approximately 100 g/mol.

Now, we can use the given information to set up a proportion to find the ratio of the actual molar mass to the molar mass of the empirical formula:

100 g/mol : 42.04 g/mol =