A compound contains 38.7% K, 13.9% N, and 47.4% O by mass. What is the empirical formula of the compound?

Answer 1

38.7% K =38.7g of K, 13.9% N= 13.9g of N, and 47.4% O =47.4g of O

38.7g of K*(1 mole of K/39.1g of K)=0.990 moles of K

13.9g of N*(1 mole of N/14.0g of N)=0.993 moles of N

47.4g of O*(1 mole of O/16.0g of O)=2.96 moles of O

Take the lowest number of moles and divide into each number of moles.

K=0.990/0.990=1

N=0.993/0.990=1

O=2.96/0.990=3

Empirical formula is KNO3

Answer 2

I left a note for you concerning your 1J calculation for physics.