Hopefully, makatulong to sa pag-aaral niyo. ❤️

Answer:

1. Na2S

% age of Na= Molar Mass of Na x Subscript of Na/Molar Mass of Na2S x 100

= 23 g of Na x 2 Numbers of Na/(23 g of Na x 2)+ (32g of S x 1) x 100

= 46 g Na/78 g Na2S x 100

= 0.589 x 100

= 58.9% of Na

% age of S = Molar Mass of S x Subscript of S/Molar Mass of Na2S x 100

= 32 g of S x 1 Number of S/(23 g of Na x 2)+ (32g of S x 1) x 100

= 32 g S/78 g x 100

= 0.410 x 100

= 41.0% of S

Total Percentage:

Na = 58.9%

S = + 41.0%

= 99.9% or 100%

2. Fe (OH)2

% age of Fe= Molar Mass of Fe x Subscript of Fe/Molar Mass of Fe(Oh)2 x 100

= 56 g of Fe x 1 Number of Fe/(56 g of Fe x 1)+(16 g of O x 2)+(1 g of H x 2) x 100

= 56 g Fe/90 g Fe(OH)2 x 100

= 0.622 x 100

= 62.2% of Fe

% age of O= Molar Mass of O x Subscript of O/Molar Mass of Fe(Oh)2 x 100

= 16 g of Fe x 2 Number/s of Fe/(56 g of Fe x 1)+(16 g of O x 2)+(1 g of H x 2) x 100

= 32 g O/90 g Fe(OH)2 x 100

= 0.355 x 100

= 35.5% of O

% age of H= Molar Mass of H x Subscript of H/Molar Mass of Fe(Oh)2 x 100

= 1 g of H x 2 Number/s of H/(56 g of Fe x 1)+(16 g of O x 2)+(1 g of H x 2) x 100

= 2 g H/90 g Fe(OH)2 x 100

= 0.022 x 100

= 2.2% of H

Total Percentage:

Fe = 62.2%

O =+ 35.5%

H =+ 2.2%

= 99.9% or 100%

3. CaSO4

% age of Ca= Molar Mass of Ca x Subscript of Ca/Molar Mass of CaSO4 x 100

= 40 g of Ca x 1 Number of Ca/(40 g of Ca x 1)+(31 g of S x 1)+(16 g of O x 4) x 100

= 40 g Ca/136 g CaSO4 x 100

= 0.294 x 100

= 29.4 % of Ca

% age of S= Molar Mass of S x Subscript of S/Molar Mass of CaSO4 x 100

= 32 g of S x 1 Number of S/(40 g of Ca x 1)+(31 g of S x 1)+(16 g of O x 4) x 100

= 32 g Ca/136 g CaSO4 x 100

= 0.235 x 100

= 23.5 % of S

% age of O= Molar Mass of O x Subscript of O/Molar Mass of CaSO4 x 100

= 16 g of O x 4 Number/s of O/(40 g of Ca x 1)+(31 g of S x 1)+(16 g of O x 4) x 100

= 64 g Ca/136 g CaSO4 x 100

= 0.470 x 100

= 47.0 % of O

Total Percentage:

Ca = 29.4%

S =+ 23.5%

O= + 47.0%

= 99.9% or 100%

4. Al (HCO3)3

% age of Al= Molar Mass of Al x Subscript of Al/Molar Mass of Al (HCO3)3 x 100

= 27 g of Al x 1 Number of Al/(27 g of Al x 1)+(1 g of H x 3)+(12 g of C x 3)+ (48g of O x 3) x 100

= 27 g Al/210 g Al (HCO3)3 x 100

= 0.128 x 100

= 12.8 % of Al

% age of H= Molar Mass of H x Subscript of H/Molar Mass of Al (HCO3)3 x 100

= 1 g of H x 3 Number/s of H/(27 g of Al x 1)+(1 g of H x 3)+(12 g of C x 3)+ (48g of O x 3) x 100

= 3 g Al/210 g Al (HCO3)3 x 100

= 0.014 x 100

= 1.4 % of H

% age of C= Molar Mass of C x Subscript of C/Molar Mass of Al (HCO3)3 x 100

= 12 g of C x 3 Number of C/(27 g of Al x 1)+(1 g of H x 3)+(12 g of C x 3)+ (48g of O x 3) x 100

= 36 g Al/210 g Al (HCO3)3 x 100

= 0.171 x 100

= 17.1 % of C

% age of O= Molar Mass of O x Subscript of O/Molar Mass of Al (HCO3)3 x 100

= (16 g of O x 3) x 3 Number/s of O/(27 g of Al x 1)+(1 g of H x 3)+(12 g of C x 3)+ (48g of O x 3) x 100

= 48 g of O x 3 Number of O/(27 g of Al x 1)+(1 g of H x 3)+(12 g of C x 3)+ (48g of O x 3) x 100

= 144 g O/210 g Al (HCO3)3 x 100

= 0.686 x 100

= 68.6 % of O

Total Percentage:

Al = 12.8%

H = + 1.4%

C = + 17.1%

O = + 68.6%

= 99.9% or 100%

5. Zn(ClO2)2

% age of Zn= Molar Mass of Zn x Subscript of Zn/Molar Mass of Zn(ClO2)2 x 100

= 65 g of Zn x 1 Number of Zn/(65 g of Zn x 1)+(35 g of Cl x 2)+(32 g of O x 2) x 100

= 65 g Zn/199 g Zn(ClO2)2 x 100

= 0.326 x 100

= 32.6 % of Zn

% age of Cl= Molar Mass of Cl x Subscript of Cl/Molar Mass of Zn(ClO2)2 x 100

= 35 g of Cl x 2 Number of Cl/(65 g of Zn x 1)+(35 g of Cl x 2)+(32 g of O x 2) x 100

= 70 g Cl/199 g Zn(ClO2)2 x 100

= 0.351 x 100

= 35.1 % of Cl

% age of O= Molar Mass of O x Subscript of O/Molar Mass of Zn(ClO2)2 x 100

= (16 g of O x 2) x 2 Number/s of O/(65 g of Zn x 1)+(35 g of Cl x 2)+(32 g of O x 2) x 100

= 32 g of O x 2 Numbers of O/(65 g of Zn x 1)+(35 g of Cl x 2)+(32 g of O x 2) x 100

= 64 g O/199 g Zn(ClO2)2 x 100

= 0.322 x 100

= 32.2 % of O

Total Percentage:

Zn = 32.6%

Cl =+ 35.1%

O =+ 32.2%

= 99.9% or 100%