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find the empirical formula of cholesterol compound that contains 83.9% Carbon, 12.0% hydrogen and 4.1% oxygen

Sagot :

Solution:

Step 1: Assume that the mass of a compound is 100 g.

[tex]\text{mass C = 83.9 g}[/tex]

[tex]\text{mass H = 12.0 g}[/tex]

[tex]\text{mass O = 4.1 g}[/tex]

Step 2: Calculate the number of moles of each element.

[tex]n \: \text{C = 83.9 g} × \frac{\text{1 mol}}{\text{12.01 g}} = \text{6.986 mol}[/tex]

[tex]n \: \text{H = 12.0 g} × \frac{\text{1 mol}}{\text{1.008 g}} = \text{11.9 mol}[/tex]

[tex]n \: \text{O = 4.1 g} × \frac{\text{1 mol}}{\text{16.00 g}} = \text{0.2562 mol}[/tex]

Step 3: Represent an empirical formula.

[tex]\text{empirical formula} = \text{C}_{x}\text{H}_{y}\text{O}_{z}[/tex]

Step 4: Divide the number of moles of each element by the least number of moles.

[tex]x = \frac{\text{6.986 mol}}{\text{0.2562 mol}} = 27[/tex]

[tex]y = \frac{\text{11.9 mol}}{\text{0.2562 mol}} = 46[/tex]

[tex]z = \frac{\text{0.2562 mol}}{\text{0.2562 mol}} = 1[/tex]

Step 5: Write the empirical formula.

[tex]\boxed{\text{empirical formula} = \text{C}_{27}\text{H}_{46}\text{O}}[/tex]

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