Question

How many cobalt atoms are in 345 g of cobalt

Answer 1

`\boxed {\boxed {\sf About \ 3.53 \ *10^(24)atoms \ Co}}`

Step-by-step 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.

`(58.93319 \ g\ Co)/(1 \ mol \ Co)`

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

`345 \ g \ Co *(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 *(1 \ mol \ Co)/(58.93319 \ g\ Co)`

`345 \ *(1 \ mol \ Co)/(58.93319 )`

`(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.

`(6.022 * 10^(23) \ atoms \ Co)/(1 \ mol \ Co)}`

Multiply this ratio by the number of moles we found.

`5.854086636 \ mol \ Co*(6.022 * 10^(23 )\ atoms \ Co)/(1 \ mol \ Co)}`

The moles of Cobalt will cancel.

`5.854086636 \ *(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.