# 300g of mg is burned on fire to produce magnesium oxide. final product weighs 4.97. what is the empirical formula for magnesium oxide?

Mar 6, 2018

On the (guessed) data we would plump for $M g O$...

#### Explanation:

The empirical formula is the simplest whole number ratio defining constituent atoms in a species... And so we interrogate the moles of magnesium and oxygen in the given problem.

$\text{Moles of magnesium} = \frac{300 \times {10}^{-} 3 \cdot g}{24.3 \cdot g \cdot m o {l}^{-} 1} = 0.0123 \cdot m o l$

$\text{Moles of oxygen} = \frac{\left(497 - 300\right) \times {10}^{-} 3 \cdot g}{16.0 \cdot g \cdot m o {l}^{-} 1} = 0.0123 \cdot m o l$

And thus there are equimolar quantities of magnesium and oxygen in the given mass so we gets...an empirical formula of $M g O$..

...to do this formally we have...$M {g}_{\frac{0.0123 \cdot m o l}{0.0123 \cdot m o l}} {O}_{\frac{0.0123 \cdot m o l}{0.0123 \cdot m o l}} \equiv M g O$

Mar 6, 2018

MgO is Magenesium Oxide. 497.5 gms

#### Explanation:

$2 M g + {O}_{2} = 2 M g O$

The molar mass of Magnesium is 24.31 gms/mol
So 300 gms will be 12.34 mols

Molar mass of MgO is 40.3 gms/mol

So 12.34 mols would be 497.5 gms
The text passage contains a chemistry problem that involves a calculation to determine the empirical formula of magnesium oxide from the given masses of magnesium and the final product. The passage also provides a detailed solution to the problem.
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Given a text passage, determine:
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First, describe the content structure you found in 1-2 sentences.

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TYPE: <one of the following values>
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system
# 300g of mg is burned on fire to produce magnesium oxide. final product weighs 4.97. what is the empirical formula for magnesium oxide?

Mar 6, 2018

On the (guessed) data we would plump for $M g O$...

#### Explanation:

The empirical formula is the simplest whole number ratio defining constituent atoms in a species... And so we interrogate the moles of magnesium and oxygen in the given problem.

$\text{Moles of magnesium} = \frac{300 \times {10}^{-} 3 \cdot g}{24.3 \cdot g \cdot m o {l}^{-} 1} = 0.0123 \cdot m o l$

$\text{Moles of oxygen} = \frac{\left(497 - 300\right) \times {10}^{-} 3 \cdot g}{16.0 \cdot g \cdot m o {l}^{-} 1} = 0.0123 \cdot m o l$

And thus there are equimolar quantities of magnesium and oxygen in the given mass so we gets...an empirical formula of $M g O$..

...to do this formally we have...$M {g}_{\frac{0.0123 \cdot m o l}{0.0123 \cdot m o l}} {O}_{\frac{0.0123 \cdot m o l}{0.0123 \cdot m o l}} \equiv M g O$

Mar 6, 2018

MgO is Magenesium Oxide. 497.5 gms

#### Explanation:

$2 M g + {O}_{2} = 2 M g O$

The molar mass of Magnesium is 24.31 gms/mol
So 300 gms will be 12.34 mols

Molar mass of MgO is 40.3 gms/mol

So 12.34 mols would be 497.5 gms
user
The text passage contains a chemistry problem that involves a calculation to determine the empirical formula of magnesium oxide from the given masses of magnesium and the final product. The passage also provides a detailed solution to the problem.

TYPE: SINGLE_QUESTION_WITH_ANSWER