Answer
a. FM of $H_{2}SO_{4}$ = 98.0 amu
b. FM of P$Cl_{5}$ = 208 amu
c. FM of N$H_{4}$Cl = 53.5 amu
d. FM of $Ca(OH)_{2}$ = 74.1 amu
Work Step by Step
a. Formula mass for $H_{2}SO_{4}$ .
FM of $H_{2}SO_{4}$ is the sum of atomic masses of all atoms in formula:
2 x AM of H = 2 x 1.00 amu = 2.00 amu
1 x AM of S = 1 x 32.06 amu = 32.06 amu
4 x AM of O = 4 x 15.99 amu = 63.96 amu
______________________________________
FM of $H_{2}SO_{4}$= 2.00 amu + 32.06 amu + 63.96 amu
FM of $H_{2}SO_{4}$= 98.02 amu
FM of $H_{2}SO_{4}$ = 98.0 amu ( to three significant figures)
b. Formula mass for P$Cl_{5}$ .
FM of P$Cl_{5}$ is the sum of atomic masses of all atoms in formula:
1 x AM P = 1 x 30.97 amu = 30.97 amu
5 x AM Cl = 5 x 35.45 amu = 177.25 amu
_______________________________
FM of P$Cl_{5}$= 30.97 amu + 177.25 amu
FM of P$Cl_{5}$ = 208.22 amu
FM of P$Cl_{5}$ = 208 amu ( to three significant figures)
c. Formula mass for ammonium chloride N$H_{4}$Cl.
FM of N$H_{4}$Cl is the sum of atomic masses of all atoms in formula:
1x AM N = 1 x 14.00 amu = 14. 00 amu
4 x AM H = 4 x 1.00 amu = 4.00 amu
1 x AM Cl = 1 x 35.45 amu = 35.45 amu
___________________________________
FM of N$H_{4}$Cl= 14. 00 amu + 4.00 amu + 35.45 amu
FM of N$H_{4}$Cl = 53.45 amu
FM of N$H_{4}$Cl = 53.5 amu ( to three significant figures)
d. Formula mass for calcium hydroxide $Ca(OH)_{2}$.
FM of $Ca(OH)_{2}$ is the sum of atomic masses of all atoms in formula:
1 x AM Ca = 1 x 40.07 amu = 40.07 amu
2 x AM O = 2 x 15.99 amu = 31.98 amu
2 x AM H = 2 x 1.00 amu = 2.00 amu
_________________________________________
FM of $Ca(OH)_{2}$ = 40.07 amu + 31.98 amu + 2.00 amu
FM of $Ca(OH)_{2}$ = 74.05 amu
FM of $Ca(OH)_{2}$ = 74.1 amu ( to three significant figures)
