Question

if 1.3860 g of mg ribbon combusts to form 2.1252 g of magnesium oxide product, calculate the experimental empirical formula from this data.

Answer 1

Answer:

Explanation:

MgO

Explanation:

The following data were obtained from the question:

mass of crucible and cover + magnesium metal = 33.741 g

mass of crucible and cover = 33.5 g

mass of crucible and cover + magnesium oxide = 33.899 g

Next, we shall determine the mass of magnesium metal. This can be obtained as follow:

mass of crucible and cover + magnesium metal = 33.741 g

mass of crucible and cover = 33.5 g

Mass of magnesium metal =..?

Mass of magnesium metal = (mass of crucible and cover + magnesium metal) – (mass of crucible and cover)

Mass of magnesium metal = 33.741 – 33.5

Mass of magnesium metal = 0.241g

Next, we shall determine the mass of magnesium oxide. This can be obtained as follow:

mass of crucible and cover + magnesium oxide = 33.899 g

mass of crucible and cover = 33.5 g

Mass of magnesium oxide =?

Mass of magnesium oxide = (mass of crucible and cover + magnesium oxide) – (mass of crucible and cover)

Mass of magnesium oxide = 33.899 –. 33.5

Mass of magnesium oxide = 0.399g

Next, we shall determine the mass of oxygen. This can be obtained as follow:

Mass of magnesium oxide = 0.399g

Mass of magnesium metal = 0.241g

Mass of oxygen =..?

Mass of oxygen = (Mass of magnesium oxide) – (Mass of magnesium metal)

Mass of oxygen = 0.399 – 0241

Mass of oxygen = 0.158g

Now, we can obtain the empirical formula for the magnesium oxide as follow:

Mg = 0.241g

O = 0.158g

Divide by their molar mass

Mg = 0.241 / 24 = 0.01

O = 0.158 / 16 = 0.0099

Divide by the smallest

Mg = 0.01 / 0.0099 = 1

O = 0.0099 / 0.0099 = 1

Therefore, the empirical formula for the magnesium oxide is MgO

Answer 2

The experimental empirical formula of the 2.1252 grams of magnesium oxide produced by the combustion of 1.3860 grams of magnesium is MgO.

To find the experimental empirical formula for the magnesium oxide product, we need to calculate the number of moles of Mg and O.

First, let's find the mass of O

$m_{Mg_{x}O_{z}} = m_{O} + m_{Mg}$

$m_{O} = m_{Mg_{x}O_{z}} - m_{Mg}$   (1)

Knowing that the mass of magnesium and magnesium oxide is 1.3860 g and 2.1252 g, respectively, we have (eq 1):

$m_{O} = m_{Mg_{x}O_{z}} - m_{Mg} = 2.1252 g - 1.3860 g = 0.7392 g$

Hence, the mass of oxygen is 0.7392 grams.

Now, we can find the number of moles of Mg and O

$n_{Mg} = \frac{m_{Mg}}{A_{Mg}} = \frac{1.3860 g}{24.305 g/mol} = 0.05702 moles$

$n_{O} = \frac{m_{O}}{A_{O}} = \frac{0.7392 g}{15.999 g/mol} = 0.04620 moles$  

Dividing the number of moles of Mg and O by the smaller number, we can find the empirical formula:

$x = \frac{0.05702}{0.04620} = 1.23 \approx 1$  

$z = \frac{0.04620}{0.04620} = 1$  

Therefore, the empirical formula of $Mg_{x}O_{z}$ is MgO.

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