Question

Nitrogen reacts with oxygen to form two compounds. compound a contains 2.8 g of nitrogen for each 1.6 g of oxygen. compound b contains 5.6 g of nitrogen for each 9.6 g of oxygen. what is the lowest whole-number mass ratio of nitrogen that combines with a given mass of oxygen?

Answer 1

Answer 1:
For compound A:
2.8 g of nitrogen for each 1.6 g of oxygen
Atomic weight of N = 14
Atomic weight of O = 16
Thus, number of moles of N = 2.8/14 = 0.2
and number of moles of O = 1.6/16 = 0.1

Thus, molar ratio of N and O in compound is 2:1.

Therefore, lowest whole-number mass ratio of nitrogen that combines with a given mass of oxygen is 2:1. And the compound formed is N2O.
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Answer 2:
For compound B:
5.6 g of nitrogen for each 9.6 g of oxygen
Atomic weight of N = 14
Atomic weight of O = 16
Thus, number of moles of N = 5.6/14 = 0.4
and number of moles of O = 9.6/16 = 0.6

Thus, molar ratio of N and O in compound is 4:6 = 2:3.

Therefore, lowest whole-number mass ratio of nitrogen that combines with a given mass of oxygen is 2:3. And the compound formed is N2O3.

Answer 2

A 2.8 / 1.6 = ____
or
b 5.6 / 9.6 = ____

(obviously neither is a whole number)

We need to assume 100g of Nitrogen.

m (2.8 / 1.6) + n (5.6 / 9.6 ) = k(100)

where:

k = integer
m and n = grams of oxygen in each compound

Solution:

m = 100-n
(100 - n) (2.8 / 1.6) + n (5.6 / 9.6) = 100k
(100 - n) 1.75 + 0.583n = 100k
-1.166n = 100k -1.75
n = -85.71k + 150

if k=1,
n = 14.3
m =85.7

thus,

one has grams O =100, grams N = 14.3 * 2.8 / 1.6 + 85.7 * 5.6 /9.8 = 25.7 + 50g N or 75g N

the ration of N/O is 3/1 or rather 3:1