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zlopas [31]
3 years ago
9

Calculate the Empirical Formula for the following compound: 0.300 mol of S and 0.900 mole of O.

Chemistry
1 answer:
ziro4ka [17]3 years ago
6 0

Answer:

\boxed {\boxed {\sf SO_3}}

Explanation:

An empirical formula shows the smallest whole-number ratio of the atoms in a compound.

So, we must calculate this ratio. Since we are given the amounts of the elements in moles, we can do this in just 2 steps.

<h3>1. Divide </h3>

The first step is division. We divide the amount of moles for both elements by the <u>smallest amount of moles</u>.

There are 0.300 moles of sulfur and 0.900 moles of oxygen. 0.300 is smaller, so we divide both amounts by 0.300

  • Sulfur: 0.300/0.300= 1
  • Oxygen: 0.900/0.300= 3

<h3>2. Write Empirical Formula</h3>

The next step is writing the formula. We use the numbers we just found as the subscripts. These numbers go after the element's symbol in the formula. Remember sulfur is S and there is 1 mole and oxygen is O and there are 3 moles.

  • S₁O₃

This formula is technically correct, but we typically remove subscripts of 1 because no subscript implies 1 representative unit.

  • SO₃

\bold {The \  empirical \ formula \ for \ the \  compound \ is  \ SO_3}}

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If you have 1.1 moles of magnesium nitrate then how many grams is that
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Answer: 162.8 grams

Explanation:

Magnesium nitrate has a chemical formula of Mg(NO3)2.

Given that:

Number of moles of Mg(NO3)2 = 1.1 moles

Mass in grams of Mg(NO3)2 = ?

For Molar mass of Mg(NO3)2, use atomic mass of magnesium = 24g, nitrogen = 14g, oxygen = 16g

Mg(NO3)2 = 24g + (14g + 16gx3) x 2

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Now, apply the formula:

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1.1 moles = Mass / 148g/mol

Mass = 1.1 moles x 148g/mol

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Thus, there are 162.8 grams of magnesium nitrate.

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Solution is here,

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