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. What is the empirical formula of ascorbic acid (vitamin C), which is 40.91% C, 4.57% H, and 54.50% 0?​

Sagot :

Solution:

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

[tex]\text{mass C = 40.91 g}[/tex]

[tex]\text{mass H = 4.57 g}[/tex]

[tex]\text{mass O = 54.50 g}[/tex]

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

[tex]n \: \text{C = 40.91 g} × \frac{\text{1 mol}}{\text{12.01 g}} = \text{3.4063 mol}[/tex]

[tex]n \: \text{H = 4.57 g} × \frac{\text{1 mol}}{\text{1.008 g}} = \text{4.5337 mol}[/tex]

[tex]n \: \text{O = 54.50 g} × \frac{\text{1 mol}}{\text{16.00 g}} = \text{3.4062 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{3.4063 mol}}{\text{3.4062 mol}} = 1 × 3 = 3[/tex]

[tex]y = \frac{\text{4.5337 mol}}{\text{3.4062 mol}} = 1.33 × 3 = 4[/tex]

[tex]z = \frac{\text{3.4062 mol}}{\text{3.4062 mol}} = 1 × 3 = 3[/tex]

Step 5: Write the empirical formula.

[tex]\boxed{\text{empirical formula} = \text{C}_{3}\text{H}_{4}\text{O}_{3}}[/tex]

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