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lilavasa [31]
2 years ago
12

A compound is 2. 00% H by mass, 32. 7% S by mass, and 65. 3% O by mass. What is its empirical formula? The final step is to use

the mole ratios of the elements to write the empirical formula. The mole ratios of the elements are: mole ratio of H:S:O = 2:1:4 The mole ratios show the ratios of the elements in the compound. Use these ratios to identify the subscripts on each element in the empirical formula. Remember that the subscript 1 is not shown. What is the correct empirical formula for this compound? HSO H2S1O4 H2SO4.
Chemistry
1 answer:
LekaFEV [45]2 years ago
3 0

The empirical formula of this compound is H_2SO_4

<u>Given the following data:</u>

  • Percentage of H = 2.00%
  • Percentage of S = 32.7%
  • Percentage of O = 65.3%

<u>Scientific data:</u>

  • Molar mass of hydrogen (H) = 1.0 g/mol.
  • Molar mass of sulfur (S) = 32 g/mol.
  • Molar mass of oxygen (O) = 16 g/mol.

To determine the empirical formula of this compound:

Note: We would assume that the mass of the compound is 100 grams.

Hence, the mass of its constituent elements are:

  • Mass of hydrogen (H) = 2.00 grams
  • Mass of sulfur (S) = 32.7 grams
  • Mass of oxygen (O) = 65.3 grams

Next, we would determine the number of moles of each element by using this formula:

Number\;of\;moles = \frac{mass}{molar\;mass}

<u>For </u><u>hydrogen</u><u> (</u><u>H</u><u>):</u>

Number\;of\;moles = \frac{2.00}{1}

Number of moles = 2.0 moles

<u>For </u><u>sulfur</u><u> (</u><u>S</u><u>):</u>

Number\;of\;moles = \frac{32.7}{32}

Number of moles = 1.0 moles

<u>For </u><u>oxygen</u><u> (</u><u>O</u><u>):</u>

Number\;of\;moles = \frac{65.3}{16}

Number of moles = 4.0 moles

Empirical formula = H_2SO_4

Read more: brainly.com/question/21280037

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Elemental analysis of caffeine yields the following results: 49.49% C, 5.15% H, 28.87% N, and the remainder is O. What is the mo
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Answer:

The answer to your question is C₄H₅N₂O

Explanation:

Process

1.- Calculate the percent of oxygen in the sample

    Percent of oxygen = 100 - 49.49 - 5.15 - 28.87

    Percent of oxygen = 16.49 %

2.- Write the percents as grams

C = 49.49 g

H = 5.15 g

N = 28.87 g

O = 16.47 g

3.- Convert the grams to moles

C          12 g ------------------- 1 mol

          49.49 g ----------------  x

             x = (49.49 x 1) 12

             x = 4.12 moles

H        1 g -------------------   1 mol

         5.15 g ----------------  x

            x = (5.15 x 1)/ 1

            x = 5.15 moles

N          14 g  --------------- 1 mol

            28.87 g ----------  x

              x = (28.87 x 1) / 14

              x = 2 mol

O           16 g ---------------- 1 mol

              16.49 g ----------- x

               x = (16.49 x 1) / 16

               x = 1.03 moles

4.- Divide by the lowest number of moles

C     4.12 /  1.03 = 4

H      5.15 / 1.03 = 5

N       2 / 1.03 = 1.9 ≈ 2

O      1.03 / 1.03 = 1

5.- Write the empirical formula

                                 C₄H₅N₂O

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