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4 years ago

C3H8 + O2 Yields CO2 + H2O help with balancing a chemical equation. need help before tomorrow.

Chemistry

2 answers:

devlian [24]4 years ago

8 0

C3H8+5O2-----3CO2+4H2O

Alik [6]4 years ago

6 0

C3H8 + 5O2 Yields 3CO2 + 4H2O

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Determine the empirical formula of a compound having the following percent composition by mass: K: 24.74%; Mn: 34.76%; O: 40.50%

Dovator [93]

Answer: The empirical formula of the compound is $KMnO_4$

Explanation:

The empirical formula is the chemical formula of the simplest ratio of the number of atoms of each element present in a compound.

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(1)

Let the mass of the compound be 100 g

Given values:

% of K = 24.7%

% of Mn = 34.76%

% of O = 40.50%

Mass of K = 24.7 g

Mass of Mn = 34.76 g

Mass of O = 40.50 g

To calculate the empirical formula of a compound, few steps need to be followed:

Step 1: Calculating the number of moles of each element

We know:

Molar mass of K = 39.10 g/mol

Molar mass of Mn = 54.94 g/mol

Molar mass of O = 16 g/mol

Putting values in equation 1, we get:

$\text{Moles of K}=\frac{24.7g}{39.10g/mol}=0.632 mol$

$\text{Moles of Mn}=\frac{34.76g}{54.94g/mol}=0.633 mol$

$\text{Moles of O}=\frac{40.50g}{16g/mol}=2.53 mol$

Step 2: Calculating the mole fraction of each element by dividing the calculated moles by the least calculated number of moles that is 0.632 moles

$\text{Mole fraction of K}=\frac{0.632}{0.632}=1$

$\text{Mole fraction of Mn}=\frac{0.633}{0.632}=1$

$\text{Mole fraction of O}=\frac{2.53}{0.632}=4$

Step 3: Writing the mole fraction as the subscripts of each of the element

The empirical formula of the compound becomes $K_1Mn_1O_4=KMnO_4$

Hence, the empirical formula of the compound is $KMnO_4$

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