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Alborosie
3 years ago
12

Calculate the percentage of each element in acetic acid, hc2h3o2, and glucose, c6h12o6.

Chemistry
1 answer:
My name is Ann [436]3 years ago
6 0
<em>Acetic acid, HC2H3O2</em>

First, calculate for the molar mass of acetic acid as shown below.
    M = 1 + 2(12) + 3(1) + 2(16) = 60 g

Then, calculating for the percentages of each element.
<em> Hydrogen:</em>
    P1 = ((4)(1)/60)(100%) = <em>6.67%</em>

<em> Carbon:</em>
   P2 = ((2)(12)/60)(100%) = <em>40%</em>

<em>Oxygen</em>
  P3 =((2)(16) / 60)(100%) = <em>53.33%</em>

<em>Glucose, C6H12O6</em>

The molar mass of glucose is as calculated below,
   6(12) + 12(1) + 6(16) = 180

The percentages of the elements are as follow,
 <em> Hydrogen:</em>
   P1 = (12/180)(100%) = <em>6.67%</em>

<em>Carbon:</em>
  P2 = ((6)(12) / 180)(100%) = <em>40%</em>

<em>Oxygen:</em>
  P3 = ((6)(16) / 180)(100%) = <em>53.33%</em>

b. Since the empirical formula of the given substances are just the same and can be written as CH2O then, the percentages of each element composing them will just be equal. 
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A compound is 52.14% C, 13.13% H, and 34.73% O. What is the empirical formula of the compound?
alina1380 [7]
% H = 100 - ( 52.14 + 34.73 )=13.13 % 

<span>assume 100 g of this compound </span>
<span>mass H = 13.13 g </span>
<span>moles H = 13.13 g / 1.008 g/mol=13 </span>

<span>mass C = 52.14 g </span>
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<span>mass O = 34.73 g </span>
<span>moles O = 34.73 g/ 15.999 g/mol=2 </span>

<span>the empirical formula is C4H13O2</span>
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