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Lapatulllka [165]
3 years ago
9

A compound contains 74.2 g Na and 25.8 g O. Determine the Empirical Formula for this compound

Chemistry
2 answers:
Ksenya-84 [330]3 years ago
4 0

Answer:

The empirical formula for the compound is Na2O

Explanation:

Data obtained from the question include:

Sodium (Na) = 74.2g

Oxygen (O) = 25.8g

We can obtain the empirical formula for the compound as follow:

First, divide the above by their individual molar mass as shown below:

Na = 74.2/23 = 3.226

O = 25.8/16 = 1.613

Next, divide the above by the smallest number

Na = 3.226/1.613 = 2

O = 1.613/1.613 = 1

Therefore, the empirical formula is:

Na2O

vazorg [7]3 years ago
3 0

Answer:

The empirical formula is Na2O

Explanation:

Step 1: Data given

Mass of Na = 74.2 grams

Molar mass of Na = 22.99 g/mol

Mass of O = 25.8 grams

Molar mass of O = 16.0 g/mol

Step 2: Calculate moles Na

Moles Na = mass Na / molar mass Na

Moles Na = 74.2 grams / 22.99 g/mol

Moles Na = 3.23 moles

Step 3: Calculate moles O

Moles O = 25.8 grams / 16.0 g/mol

Moles O = 1.61 moles

Step 4: Calculate the mol ratio

We divide by the smallest amount of moles

Na: 3.23 moles / 1.61 moles = 2

O: 1.61 moles / 1.61 moles = 1

This means for every O atom we have 2 Na atoms

The empirical formula is Na2O

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14.7 grams of magnesium reacts completely with 9.7 grams of oxygen to form magnesium oxide (MgO). What is the percent compositio
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The answer is 60.3% magnesium, 39.7% oxygen.
Solution:
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Since magnesium reacts completely with oxygen, it is the limiting reactant in the reaction. Hence, we can use the number of moles of magnesium to get the mass of MgO produced:
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We also use the number of moles of magnesium to get the mass of oxygen consumed in the reaction:
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Notice that we can just subtract the magnesium's percentage from 100% to get 
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