Question

What is the mass of 1.5 x 10^21 silicon atoms?

Answer 1

0. 070g

Mass of 1 Si atom =28.08/6.022*1023

So mass 1.5 *1021*4.6629*10-23

Answer 2

`\boxed {\boxed {\sf 0.070 \ g \ Si}}`

Step-by-step explanation:

We are given a value of atoms and asked to find the mass. We can complete this in 2 steps.

1. Convert Atoms to Moles

First, we have to convert atoms to moles. We use Avogadro's Number or 6.022 × 10²³. This is the number of particles (atoms, molecules, formula units, etc.) in 1 mole of a substance.

In this case, the particles are silicon atoms. We can set up a proportion.

`\frac { 6.022 * 10^(23) \ atom \ Si }{1 \ mol \ Si}`

Multiply by the value of atoms given.

`1.5 * 10^(21) \ atom \ Si *\frac { 6.022 * 10^(23) \ atom \ Si }{1 \ mol \ Si}`

Flip the ratio. This allows the atoms of silicon units to cancel and leave the moles of silicon.

`1.5 * 10^(21) \ atom \ Si *\frac {1 \ mol \ Si } {6.022 * 10^(23) \ atom \ Si}`

`1.5 * 10^(21)*\frac {1 \ mol \ Si } {6.022 * 10^(23) }`

`\frac {1.5 * 10^(21) \ mol \ Si } {6.022 * 10^(23) }`

`0.002490866822 \ mol \ Si`

2. Convert Moles to Grams

Next, we convert the moles to grams. We use the molar mass, which is found on the Periodic Table. This value is the number of grams in 1 mole of a substance. Look for silicon (Si).

• Si: 28.085 g/mol

Use this value as a ratio.

`\frac {28.085 \ g \ Si}{1 \ mol \ Si}`

Multiply by the number of moles we calculated.

`0.002490866822 \ mol \ Si *\frac {28.085 \ g \ Si}{1 \ mol \ Si}`

The moles of silicon units cancel, so we are left with grams.

`0.002490866822 *\frac {28.085 \ g \ Si}{1}`

`0.002490866822 *{28.085 \ g \ Si}`

`0.06995599469 \ g \ Si`

3. Round

Finally we must round our answer. The original value has 2 significant figures, so our answer must have the same. For the number we calculated, that is the thousandth place.

`0.069\underline95599469 \ g Si`

The 9 tells us to round the other 9 to a 0, then the 6 to a 7.

`0.070 \ g \ Si`

The mass of 1.5 × 10²¹ silicon atoms is 0.070 grams of silicon.