If 1.9022g pet is burned in oxygen it produces 0.6585 g h2o and 4.0216 g Co2. The empirical formula of PET is [tex]C_{10}[/tex][tex]H_{8}[/tex][tex]O_{5}[/tex].
We are told that 1.250 g of PET is burned to produce 0.4328 g [tex]H_{2}[/tex]O and 2.643g C [tex]O_{2}[/tex].
By using percentage composition we can determine the masses of carbon dioxide and hydrogen from pet.
% composition = [tex]\frac{molar mass of element in compound }{molar mass of compound }[/tex] x 100%
Now for Carbon.
% composition C = [tex]\frac{12.01 g/mole}{12.01 g/mole + (2 x 16.00 g/mol)}[/tex] x 100% = 27.29%
Now multiplying the mass of carbon dioxide by percentage composition we can get the mass of C.
2.643g x 27.29% = 0.7213g
% composition H = [tex]\frac{2 x 1.01 g/mol }{16.00 g/mol + ( 2 x 1.01 g/mol)}[/tex] x 100%
Now multiplying the mass of water by percentage composition of hydrogen we can get the mass of H.
0.4328 g × 11.21 % = 0.04852 g
By subtracting the mass of carbon and hydrogen from the mass of the PET i.e. 1.250 g we can get the mass of oxygen in PET.
1.250 g − 0.04852 g − 0.7213 g = 0.4802 g
Let's calculate the moles of each component.
Moles of carbon = 0.7213/ 12.01 g/mol = 0.06006 mol
Moles of hydrogen = 0.0485/ 1.01 g/mol = 0.04804 mol
Moles of oxygen = 0.4802/ 16.00 g/mol = 0.03001 mol
Now, to get the subscripts for the empirical formula, lets divide the moles of each element with the lowest number of moles.
Carbon = 0.06006/0.03001 = 2
Hydrogen = 0.04804/0.03001 = 1.6
Oxygen = 0.03001/0.03001 = 1
We see that 1.6 is not a whole integer and is equivalent to the fraction 8/5. So the empirical formula become [tex]C_{2}[/tex][tex]H\frac{8}{5}[/tex]O which seems inappropriate as our empirical formulas do not include fractions. So, to fix this problem, let's multiply all the subscripts with the fraction 5.
The final empirical formula is [tex]C_{10}[/tex][tex]H_{8}[/tex][tex]O_{5}[/tex].
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