A 1.344g sample of a compound contains 0.365g Na, 0.221g N, and 0.758g O.What is its empirical formula?

Answer 1

first we have to find the number of moles of each element. To do this we divide it by the molar mass of each.
Na = 0.365/23
=0.01587
N = 0.221/14
= 0.001579
O = 0.758/16
= 0.04738

Now divide these by the smallest number i.e. 0.01579
To get,
Na = 1
N = 1
O =3
SO the empirircal formula will be = NaNO3

Answer 2

Consider these facts:

-Atomic masses (rounded up) of :
-Na: 23 g/mol
-N: 14 g/mol
-O: 16 g/mol

Now, divide each amount by it's corresponding atomic mass, like this:

0.365g Na / 23g = 0.015g
0.221g N / 14g = 0.015g
0.758g O / 16g = 0.047g

And finally, divide each result by the smallest number, which in this case will be 0.015:

Na--> 0.015 / 0.015 = 1
N--> 0.015 / 0.015 = 1
O--> 0.047 / 0.015= 3.1 -->3 (rounded down)

So, you end up with:

Na: 1
N:1
O:3

Empirical formula: NaNO3.

Hope this is useful.

Answer 3

Firstly, find the percentage of each of the elements in the sample
%Na = 0.365/1.344 x 100
=27.158%
%N=0.221/1.344x100
=16.44%
%O=0.758/1.344x100
56.4%

Convert to mass (in 100g)
Na=27.158g
N=16.44g
O=56.4g

Change to moles (n = m/M)
n(Na) = 27.158/22.99
=1.18mol
n(N) = 16.44/14.01
=1.173mol
n(O) = 56.4/16
=3.525mol

Find the simplest ratio. (The element with the smallest amount of moles over the moles of the other elements.)
N is the smallest.
Ratio of Na = 1.18/1.173
=1
Ratio of N = 1.173/1.173
=1
Ratio of O = 3.525/1.173
=3

Therefore the empirical formula is Na1N1O3 = NaNO3