Determine the number of atoms and the mass of zirconium, silicon and oxygen found in 0.3384 mol of zircon zrsio4, a
semiprecious stone.

Explanation:

1 mol of ZrSiO₄ , have , 1 mol of Zr , 1 mol of Si and  4 mol of  Oxygen .

since ,

1 mol of a substance have 6.022 * 10 ²³ particles ,

From the question ,

0.3384 mol of zircon ZrSiO₄ ,

hence ,

from the above equation ,

1 mol of a substance have 6.022 * 10 ²³ particles ,

0.3384 mol = 0.3384 * 6.022 * 10 ²³ number of atoms

                  = 2.037 * 10 ²³ number of atoms

Hence ,

0.3384 mol of ZrSiO₄ will have , 0.3384 mol of Zr ,0.3384 mol of Si and  4*0.3384 mol of  Oxygen .

Hence ,

The mass is calculated as -

since ,

Moles is denoted by given mass divided by the molecular mass ,  

Hence ,  

n = w / m

n = moles ,  

w = given mass ,  

m = molecular mass .

n = w / m

w = n * m

Since we know , the molecular mass of Zr = 91 g/mol

moles of Zr = 0.3384 mol

using the above formula ,

w = n * m

w = 0.3384 mol  * 91 g/mol = 30.79 g

hence ,

The mass of Zr = 30.79 g

Since we know , the molecular mass of Si = 28 g/mol

moles of Si = 0.3384 mol

using the above formula ,

w = n * m

w = 0.3384 mol  * 28 g/mol = 9.475 g

hence ,

The mass of Si = 9.475 g

Since we know , the molecular mass of O = 16 g/mol

moles of Si = 1.3536 mol

using the above formula ,

w = n * m

w = 1.3536 mol  * 16 g/mol = 21.65 g

hence ,  

The mass of O = 21.65 g

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