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Keith_Richards [23]
3 years ago
8

What are the empirical formula and empirical formula mass for C10H30O10?

Chemistry
1 answer:
AysviL [449]3 years ago
4 0

Answer:

Empirical formula: CH₃O

Empirical formula mass = 31 g/mol

Explanation:

Data Given:

Molecular Formula = C₁₀H₃₀O₁₀

Empirical Formula = ?

Empirical Formula mass =

Solution

Empirical Formula:

Empirical formula is the simplest ration of atoms in the molecule but not all numbers of atoms in a compound.

So,

The ratio of the molecular formula should be divided by whole number to get the simplest ratio of molecule

As

C₁₀H₃₀O₁₀ Consist of  10 Carbon (C) atoms, 30 Hydrogen (H) atoms, and 10 Oxygen (O) atoms.

Now

Look at the ratio of these three atoms in the compound

                         C : H : O

                        10 : 30 : 10

Divide the ratio by two to get simplest ratio

                          C      :   H      :    O

                         10/10 : 30/10 : 10/10

                             1 : 3 : 1

So for the empirical formula the simplest ratio of carbon to hydrogen to oxygen is 1:3:1

So the empirical formula will be

                     Empirical formula of C₁₀H₃₀O₁₀ =  CH₃O

Now

To find the empirical formula mass in g/mol

Formula mass:

Formula mass is the total sum of the atomic masses of all the atoms present in a formula unit.

**Note:

if we represent the molar mass of the empirical formula for one mol in grams then it is written as g/mol

So,

As the empirical formula of C₁₀H₃₀O₁₀ is CH₃O

Then Its empirical formula mass will be

CH₃O

Atomic Mass of C = 12

Atomic Mass of H = 3

Atomic Mass of O = 16

Total Molar mass of CH₃O

CH₃O = 12 + 3(1) + 16

CH₃O = 12 + 3 + 16

CH₃O = 31 g/mol

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The thermal decomposition of calcium carbonate will produce 14 g of calcium oxide. The stoichiometric ratio of calcium carbonate to calcium oxide is 1:1, therefore the number of moles of calcium carbonate decomposed is equal to the number of moles of calcium oxide formed.

Further Explanation:

To solve this problem, follow the steps below:

  1. Write the balanced chemical equation for the given reaction.
  2. Convert the mass of calcium carbonate into moles.
  3. Determine the number of moles of calcium oxide formed by using the stoichiometric ratio for calcium oxide and calcium carbonate based on the coefficient of the chemical equation.
  4. Convert the number of moles of calcium oxide into mass.

Solving the given problem using the steps above:

STEP 1: The balanced chemical equation for the given reaction is:

CaCO_{3} \rightarrow \ CaO \ + \ CO_{2}

STEP 2: Convert the mass of calcium carbonate into moles using the molar mass of calcium carbonate.

mol \ CaCO_{3} \ = 25 \ g \ CaCO_{3} \ (\frac{1 \ mol \ CaCO_{3}}{100.0869 \ g \ CaCO_{3}})\\ \\\boxed {mol \ CaCO_{3} \ = 0.2498 \ mol}

STEP 3: Use the stoichiometric ratio to determine the number of moles of CaO formed.

For every mole of calcium carbonate decomposed, one more of a calcium oxide is formed. Therefore,

mol \ CaO \ = 0.2498 \ mol

STEP 4: Convert the moles of CaO into mass of CaO using its molar mass.

mass \ CaO \ = 0.2498 \ mol \ CaO \ (\frac{56.0774 \ g \ CaO}{1 \ mol \ CaO})\\ \\mass \ CaO \ = 14.008 \ g

Since there are only 2 significant figures in the given, the final answer must have the same number of significant figures.

Therefore,

\boxed {mass \ CaO \ = 14 \ g}

Learn More

  1. Learn more about stoichiometry brainly.com/question/12979299
  2. Learn more about mole conversion brainly.com/question/12972204
  3. Learn more about limiting reactants brainly.com/question/12979491

Keywords: thermal decomposition, stoichiometry

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