We can write the reaction for this problem as:
`~N_2` + `~O_2` -> `~N_xO_y`
x = subscript for N in the product
y = subscript for O in the product
Determine the grams of `~O_2 ` :
We are given the grams of `~N_2` and `~N_xO_y` :
`~N_2` = 12.04 g
`~N_xO_y` = 39.54 g
We can use the Law of Conservation of Mass to determine the grams of `~O_2` .
According to the Law of Conservation of Mass, the mass of the reactants in a chemical reaction is equal to the mass of the products.
mass `~N_2` + mass `~O_2` = mass `~N_xO_y`
12.04 g + ? g `~O_2` = 39.54 g
` ~O_2` = 39.54 g – 12.04 g
= 27.5 g``
Determine the empirical formula:
Step 1: Convert grams of nitrogen and oxygen to moles.
Convert grams to moles by dividing by the molar mass of each element.
N: 12.04 g x (1 mol/14.007 g) = 0.860 mol
O: 27.5 g x (1 mol/15.999 g) = 1.72 mol
Step 2: Divide both answers from Step 1 by the smallest answer from Step 1.
N: 0.860/0.860 = 1
O: 1.72/0.860 = 2
Step 3: Use the answers to Step 2 as subscripts for the empirical formula.
The empirical formula is `~NO_2` ` `
Determine the molecular formula:
We are given the molar mass of the product `~N_xO_y` as 92.02 g
Step 1: Determine the molar mass of the empirical formula `~NO_2` .
molar mass of `~NO_2 ` = 14.007 + (2)(15.999) = 46.005 g
Step 2: Divide the given molar mass of the product by the molar mass of the empirical formula.
92.02/46.005 = 2
Step 3: Multiply the subscripts of the empirical formula by the answer to Step 2.
2 x `~N_1O_2` = `~N_2O_4`
The molecular formula is `~N_2O_4` .
Determine the percent composition of the product `~N_2O_4` .
Step 1: Determine the molar mass of each element in the product.
N: (2)(14.007) = 28.014 g
O: (4)(15.999) = 63.996 g
Step 2: Determine the molar mass for the entire product.
`~N_2O_4` : 28.014 + 63.996 = 92.01 g
Step 3: Divide the molar mass of each element by the molar mass of the entire product and multiply by 100.
% N: (28.014/92.01) x 100 = 30.4%
% O: (63.996/92.01) x 100 = 69.6%
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