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Oxana [17]
2 years ago
15

The main constituent of gallstones is cholesterol. Cholesterol may have a role in heart attacks and blood clot formation. Its el

emental percentage composition is 83.87% C, 11.99% H, and 4.14% O. It has a molecular weight of 386.64 amu. Calculate its empirical and molecular formulas.
Chemistry
1 answer:
Nataly [62]2 years ago
7 0

The main constituent of gallstones is cholesterol. Cholesterol may have a role in heart attacks and blood clot formation. Its elemental percentage composition is 83.87% C, 11.99% H, and 4.14% O. It has a molecular weight of 386.64 amu. Empirical formula is C₃H₄O₁ and Molecular formula is 7(C₃H₄O₁).

<h3>What is Empirical Formula ?</h3>

Empirical formula is the simplest whole number ratio of atoms present in given compound.

Element   %   Atomic mass   Relative no. of atoms  Simplest whole ratio

C            83.87       12                     \frac{83.87}{12} = 6.98                   \frac{6.98}{0.25} = 3

H            11.99         1                       \frac{11.99}{1} = 11.09                  \frac{11.09}{0.25} = 4

O            4.14          16                      \frac{4.14}{16} = 0.25                    \frac{0.25}{0.25} = 1

Thus the empirical formula is C₃H₄O₁.

<h3>How to find the Molecular formula of compound ?</h3>

Molecular formula = Empirical formula × n

n = \frac{\text{Molecular weight}}{\text{Empirical formula weight}}

  = \frac{386.64}{56}

  = 7

Molecular formula = Empirical formula × n

                               = 7 (C₃H₄O₁)

Thus from the above conclusion we can say that The main constituent of gallstones is cholesterol. Cholesterol may have a role in heart attacks and blood clot formation. Its elemental percentage composition is 83.87% C, 11.99% H, and 4.14% O. It has a molecular weight of 386.64 amu. Empirical formula is C₃H₄O₁ and Molecular formula is 7(C₃H₄O₁).

Learn more about the Empirical Formula here: brainly.com/question/1603500

#SPJ4

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Answer:

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Explanation:

There is some missing info. I think this is the complete question.

<em>A student dissolves 4.6 g of glucose in 500 mL of a solvent with a density of 0.87 g/mL. The student notices that the volume of the solvent does not change when the glucose dissolves in it. Calculate the molarity and molality of the student's solution. Round both of your answers to 2 significant digits.</em>

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The molar mass of glucose is 180.16 g/mol.

4.6 g × 1 mol/180.16 g = 0.026 mol

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