Question

A. How many ATOMS of silicon are present in 5.53 grams of silicon dioxide?

Answer 1

5.55 x 10²² atoms of silicon (Si).

Step-by-step explanation:

Measurements => Dimensional Analysis.

The Mole => Avogadro's Number.

Remember that we use units of measurement to solve problems.

The Avogadro's number is 6.022 x 10²³ atoms/mol. This is telling us that we have 6.022 x 10²³ atoms of an element in 1 mol.

In this case, we have 5.53 g of silicon dioxide (SiO2). The molar mass of SiO2 can be calculated by using the periodic table which is 60 g/mol. We convert from mass of SiO2 to moles. You can see that 1 mol of SiO2 contains just 1 mol of silicon (Si). When we found the number of moles of Si, we use the Avogadro's number to find the number of atoms that Si has. This process known as a dimensional analysis will look like this:

`5.53\text{ g SiO}_2\cdot\frac{1\text{ mol SiO}_2}{60\text{ g SiO}_2}\cdot\frac{1\text{ mol Si}}{1\text{ mol SiO}_2}\cdot\frac{6.022\cdot10^(23)\text{ atoms}}{1\text{ mol}}=5.55\cdot10^(22)\text{ atoms Si.}`

The answer would be that we have 5.55 x 10²² atoms of silicon (Si) in 5.53 g of silicon dioxide (SiO2).