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What is the percent composition of the following

*KBr
*LiOH
*Mgl2
*Al2(SO4)3

User Cyriel
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1 Answer

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Final answer:

The percent composition of a compound can be calculated by dividing the mass of each element by the molar mass of the compound and multiplying by 100. For example, the percent composition of KBr is 32.77% potassium and 67.23% bromine. Similarly, the percent composition of LiOH is 28.95% lithium, 66.81% oxygen, and 4.23% hydrogen.

Step-by-step explanation:

The percent composition of a compound is the percent by mass of each element in the compound. To calculate percent composition, we divide the mass of each element by the molar mass of the compound and multiply by 100. Let's calculate the percent composition for each compound:

a) KBr:

Atomic mass of K = 39.10 g/mol

Atomic mass of Br = 79.90 g/mol

Molar mass of KBr = 39.10 + 79.90 = 119 g/mol

Percent composition of K in KBr = (39.10 / 119) * 100 = 32.77%

Percent composition of Br in KBr = (79.90 / 119) * 100 = 67.23%

Therefore, the percent composition of KBr is 32.77% potassium and 67.23% bromine.

b) LiOH:

Atomic mass of Li = 6.94 g/mol

Atomic mass of O = 16.00 g/mol

Atomic mass of H = 1.01 g/mol

Molar mass of LiOH = 6.94 + 16.00 + 1.01 = 23.95 g/mol

Percent composition of Li in LiOH = (6.94 / 23.95) * 100 = 28.95%

Percent composition of O in LiOH = (16.00 / 23.95) * 100 = 66.81%

Percent composition of H in LiOH = (1.01 / 23.95) * 100 = 4.23%

Therefore, the percent composition of LiOH is 28.95% lithium, 66.81% oxygen, and 4.23% hydrogen.

c)MgCl₂ :

Atomic mass of Mg = 24.31 g/mol

Atomic mass of Cl = 35.45 g/mol

Molar mass of MgCl₂ = 24.31 + (2 * 35.45) = 95.21 g/mol

Percent composition of Mg in MgCl₂ = (24.31 / 95.21) * 100 = 25.55%

Percent composition of Cl in MgCl₂ = (2 * 35.45 / 95.21) * 100 = 74.45%

Therefore, the percent composition of MgCl₂ is 25.55% magnesium and 74.45% chlorine.

d) Al₂(SO₄)₃ :

Atomic mass of Al = 26.98 g/mol

Atomic mass of S = 32.07 g/mol

Atomic mass of O = 16.00 g/mol

Molar mass of Al₂(SO₄)₃ = (2 * 26.98) + (3 * 32.07) + (12 * 16.00) = 342.15 g/mol

Percent composition of Al in Al₂(SO₄)₃ = (2 * 26.98 / 342.15) * 100 = 15.78%

Percent composition of S in Al₂(SO₄)₃ = (3 * 32.07 / 342.15) * 100 = 28.14%

Percent composition of O in Al₂(SO₄)₃ = (12 * 16.00 / 342.15) * 100 = 56.08%

Therefore, the percent composition of Al₂(SO₄)₃ is 15.78% aluminum, 28.14% sulfur, and 56.08% oxygen.

User Pancakeo
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