What is the percent composition of nitrogen in a 2.57 g sample of Al(NO3)3?

19.8% of Nitrogen

Explanation:

In the Al(NO₃)₃ there are:

1 atom of Al

3 atoms of N

And 9 atoms of O

The molar mass of Al(NO₃)₃ is:

1 Al * (26.98g/mol) = 26.98g/mol

3 N * (14g/mol) = 42g/mol

9 O * (16g/mol) = 144g/mol

26.98 + 42 + 144 = 212.98g/mol

We can do a conversion using these molar masses to find the mass of nitrogen is the sample, that is:

2.57g * (42g/mol / 212.98g/mol) =

0.51g N

Percent composition of nitrogen is:

0.51g N / 2.57g * 100

= 19.8% of Nitrogen

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the atomic number of carbon is six, which means that a carbon atom has six protons. carbon has three naturally occurring isotopes: carbon-12, carbon-13, and carbon-14. which of these statements are true about carbon and its isotopes? select all that appl

a. the law of proportion states the combination of 2 atoms in a specific ratio to produce multiple compounds such as the given statement of question ‘1st compound contains 88.8 % cu ’ indicates 88.8 g of cu is present along with 11.2 g oxygen in 100 g of compound whereas the statement ‘2st compound contains 79.9 % cu ’ indicates 79.9 g of cu is present along with 20.1 g oxygen in 100 g of compound

1st compound contains 88.8 g cu

for cu

m = 63 g /mol

atomic % = 88.8 /63 = 1.42

for o

m = 16 g /mol

atomic % = 11.2 /16 = 0.7

b. the emperical formula of 1st compound is

cu1.4o0.7

cu2o