Asked by Anonymous
what is the empirical formula for a compund with 26.2 g N, 7.5 g H, and 66.3 g Cl?
Answers
Answered by
Devron
26.2 g of N*(1 mole of N/14.01g)= moles of N=1.87 moles of N
7.5 g of H *(1 mole of H/1.007 g of H)= moles of H=7.45 moles of H
66.3 g of Cl *(1 mole of Cl/35.45g of Cl)= moles of Cl=1.87 moles of Cl
take the lowest number of moles, and divide by the others.
N=1.87/1.87=1
H=7.45/1.87=3.98=4
Cl=1.87/1.87=1
7.5 g of H *(1 mole of H/1.007 g of H)= moles of H=7.45 moles of H
66.3 g of Cl *(1 mole of Cl/35.45g of Cl)= moles of Cl=1.87 moles of Cl
take the lowest number of moles, and divide by the others.
N=1.87/1.87=1
H=7.45/1.87=3.98=4
Cl=1.87/1.87=1