How many grams are in 9.03×1023atoms of Copper (Cu)?

9.03 × 10²³ atoms of copper have a mass of 95.3 g.

We want to convert 9.03 × 10²³ atoms of copper to grams. We will need a series of conversion factors.

A conversion factor is an arithmetical multiplier for converting a quantity expressed in one set of units into an equivalent expressed in another.

The required conversion factor is Avogadro's number: there are 6.02 × 10²³ atoms in 1 mole of atoms.

9.03 × 10²³ atoms × (1 mol/6.02 × 10²³ atoms) = 1.50 mol

The required conversion factor is the molar mass of copper: 63.55 g/mol.

1.50 mol × 63.55 g/mol = 95.3 g

9.03 × 10²³ atoms of copper have a mass of 95.3 g.

Learn more about Avogadro's number here: link

Mass of Cu : 95.319 g

Given

9.03 x 10²³atoms of Copper

Required

mass of Cu

Solution

The mole is the number of particles(molecules, atoms, ions) contained in a substance  

1 mol = 6.02.10²³ particles

Can be formulated

N=n x No

N = number of particles

n = mol

No = Avogadro's = 6.02.10²³

Mol of Copper :

n = N/No

n = 9.03 x 10²³/6.02.10²³

n = 1.5

Mass of Copper :

= mol x Ar Cu

= 1.5 x 63,546 g/mol

= 95.319 g

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