Penjelasan:

1. Menentukan % massa = ( jumlah atom x Ar ) / Mr senyawa x 100%

Mr CaCO3 = 100

massa Ca = ( 1 x 40 ) / 100 x 100% = 40%

massa C = ( 1 x 12 ) / 100 x 100% = 12%

massa O = ( 3 x 16 ) / 100 x 100% = 48%

2. berat unsur H = ( 2 x Ar H ) / Mr H2O x massa H2O

= ( 2 x 1 ) / 18 x 27

= 3 gram

berat unsur O = ( 1 x Ar O ) / Mr H2O x massa H2O

= ( 1 x 16 ) / 18 x 27

= 24 gram

3. a. Rumus empiris

mol C = massa C / Ar C = 90% / 12 = 0,075 mol

mol H = massa H / Ar H = 10% / 1 = 0,1 mol

Perbandingan mol C : H = 0,075 : 0,1 = 3 : 4

Jadi rumus empirisnya C3H4

b. Rumus empiris x n = Mr

( C3H4 ) n = 80

( 3x Ar C + 4x Ar H ) n = 80

( 3 x12 + 4x1 ) n = 80

40 n = 80 => n = 2

Rumus molekul ( C3H4 ) 2 => C6H8

4. massa O = 23 - 12 - 3 = 8 gram

Rumus empiris

mol C = 12 / 12 = 1 mol

mol H = 3 / 1 = 3 mol

mol O = 8 / 16 = 0,5 mol

Perbandingan mol C : H : O = 1 : 3 : 0,5 = 2 : 6 : 1

Jadi rumus empirisnya C2H6O

5. Jumlah volume = jumlah koefisien dalam reaksi

10 PxOy -> 5 P4 + 15 O2

10 Px = 20 P => x = 2

10 Oy = 30 O => y = 3

Maka rumusnya P2O3