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3 years ago

What is the empirical formula for a compound if a sample contains 1.0 g of s and 1.5 g of o?

1 answer:

4 0

The key principle here is to find the mole ratio of each element to know the empirical formula. First we need to convert the mass to moles by using the molar mass of each element. The molar mass of S is 32 and O is 16 grams/mole. The moles of S and O given the mass is 0.03125 and 0.09375 moles, respectively. The compound would need 3 moles of O per S. the compound is SO3

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3 years ago

Which of the following is an essential condition for a combustion reaction ?

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4 years ago

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Viefleur [7K]

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3 years ago

4. Magnesium and oxygen undergo a chemical reaction to

erastovalidia [21]

Answer:

About 16.1 grams of oxygen gas.

Explanation:

The reaction between magnesium and oxygen can be described by the equation:
$\displaystyle 2\text{Mg} + \text{O$_2$} \longrightarrow 2\text{MgO}$

24.4 grams of Mg reacted with O₂ to produce 40.5 grams of MgO. We want to determine the mass of O₂ in the chemical change.

Compute using stoichiometry. From the equation, we know that two moles of MgO is produced from every one mole of O₂. Therefore, we can:

Convert grams of MgO to moles of MgO.
Moles of MgO to moles of O₂
And moles of O₂ to grams of O₂.

The molecular weights of MgO and O₂ are 40.31 g/mol and 32.00 g/mol, respectively.

Dimensional analysis:

$\displaystyle 40.5\text{ g MgO} \cdot \frac{1\text{ mol MgO}}{40.31\text{ g MgO}} \cdot \frac{1\text{ mol O$_2$}}{2\text{ mol MgO}} \cdot \frac{32.00\text{ g O$_2$}}{1\text{ mol O$_2$}} = 16.1\text{ g O$_2$}$

In conclusion, about 16.1 grams of oxygen gas was reacted.

You will obtain the same result if you compute with the 24.4 grams of Mg instead:

$\displaystyle 24.4\text{ g Mg}\cdot \frac{1\text{ mol Mg}}{24.31\text{ g Mg}} \cdot \frac{1\text{ mol O$_2$}}{1\text{ mol Mg}} \cdot \frac{32.00\text{ g O$_2$}}{1\text{ mol O$_2$}} = 16.1\text{ g O$_2$}$

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2 years ago