Part A 1.0 x 10^23 gold atoms Express your answer using two significant figures. m= Part B ...

Question:

Part A

1.0 x 10^23 gold atoms

Express your answer using two significant figures.

m=

Part B

2.90 x 10^22 helium atoms

m=

Mass and Number of Atoms:

Expressing amounts of elements or compounds is a special application of molar mass, an essential topic in chemistry. Problems such as relating these amounts are solved typically using the molar mass/molecular weight of the specific element of compound. When converting from atoms to mass (and vice versa), the molar mass and the Avogadro's number are used as conversion ratios that are multiplied to the given amount to obtain the desired quantity.

Part A

The mass of gold (Au) given {eq}\rm 1.0 \times 10^{23}\ Au\ atoms {/eq} is obtained using the Avogadro's number and the molar mass of gold,

$$\rm MM_{Au} = \rm 197\ g\ Au$$.

So,

$$\rm mass\ of\ Au = \rm 1.0 \times 10^{23}\ Au\ atoms \times \frac{1\ mol\ Au}{6.022 \times 10^{23}\ Au\ atoms} \times \frac{197\ g\ Au}{1\ mol\ Au} = \rm 33\ g\ Au$$

Hence, in {eq}\rm 1.0 \times 10^{23}\ Au\ atoms {/eq}, the mass is

$$\boxed{\rm m = \rm 33\ g\ Au}$$

{eq}\\ {/eq}

Part B

The molar mass of Helium (He) is

$$\rm MM_{He} = \rm 4.00\ g\ He$$

With this,

$$\rm mass\ of\ He = \rm 2.90 \times 10^{22}\ He\ atoms \times \frac{1\ mol\ He}{6.022 \times 10^{23}\ He\ atoms} \times \frac{4.00\ g\ He}{1\ mol\ He} = \rm 0.19\ g\ He$$

Therefore, the mass of He in {eq}\rm 2.90 \times 10^{22}\ He\ atoms {/eq} is

$$\boxed{\rm m = \rm 0.19\ g\ He}$$ 