Question

How many cobalt atoms are in 345 g of cobalt

Answer 1

Answer:

$\boxed {\boxed {\sf About \ 3.53 \ *10^{24}atoms \ Co}}$

Explanation:

To convert from grams to atoms:

1. Convert grams to moles
2. Convert moles to grams

1. Convert grams to moles

First, find the molar mass of cobalt using the Periodic Table of Elements.

• Cobalt (Co): 58.93319 g/mol

Next, use this mass as a ratio or fraction.

$\frac{58.93319 \ g\ Co}{1 \ mol \ Co}$

Multiply the mass of the given sample (345 grams) by this ratio.

$345 \ g \ Co *\frac{58.93319 \ g\ Co}{1 \ mol \ Co}$

Flip the fraction so the grams of Cobalt will cancel each other out when multiplying.

$345 \ g \ Co *\frac{1 \ mol \ Co}{58.93319 \ g\ Co}$

$345 \ *\frac{1 \ mol \ Co}{58.93319 }$

$\frac{345 \ mol \ Co}{58.93319 } =5.854086636 \ mol \ Co$

2. Convert moles to atoms

Use Avogadro's Number, which tells us the number of units (in this case atoms) in 1 mole.

• $6.022 *10^{23} \ atoms/mol$

Use this number as a ratio or fraction.

$\frac{6.022 * 10^{23} \ atoms \ Co}{1 \ mol \ Co}}$

Multiply this ratio by the number of moles we found.

$5.854086636 \ mol \ Co*\frac{6.022 * 10^{23 }\ atoms \ Co}{1 \ mol \ Co}}$

The moles of Cobalt will cancel.

$5.854086636 \ *\frac{6.022 * 10^{23 }\ atoms \ Co}{1 }}$

$5.854086636 \ *{6.022 * 10^{23 }\ atoms \ Co}$

$3.52533097*10^{24} \ atoms \ Co$

3.Round

The original measurement of 345 has 3 significant figures (3, 4, and 5). We must round to 3 sig figs, which is the hundredth place for this measurement.

$3.52533097*10^{24} \ atoms \ Co$

The 5 in the thousandth place tells us to round the 2 to a 3.

$3.53 \ *10^{24}atoms \ Co$

There are about 3.53*10²⁴ atoms of cobalt in 345 grams.