What is the mass in grams of #9.357 * 10^30# atoms of iron?

Answer 1

#8.677 * 10^(8)"g"#

An interesting approach to use here is to use iron's molar mass and Avogadro's number to find the mass of a single atom of iron, which you can then be used as a conversion factor to find the mass of #9.357 * 10^(30)# atoms.
#"mass of one mole " stackrel(color(red)(1)color(white)(aa))(->) " mass of one atom " stackrel(color(purple)(2)color(white)(aa))(->) " mass of"color(white)(a) 9.357 * 10^(30)"atoms"#
#color(white)()# #color(red)(1.)#
The mass of one mole of any element is given by its molar mass. Iron has a molar mass of #"55.845 g mol"^(-1)#, which means that every mole of iron has mass of #"55.845 g"#.
Now, one mole of any element is defined as #6.022 * 10^(23)# atoms of that element -- this is known as Avogadro's number.
#color(blue)(|bar(ul(color(white)(a/a)"1 mole" = 6.022 * 10^(23)"atoms"color(white)(a/a)|)))#
So, if one mole of iron contains #6.022 * 10^(23)# atoms of iron, and if the mass of one mole if #"55.845 g"#, it follows that the mass of one atom of iron will be
#1 color(red)(cancel(color(black)("atom Fe"))) * "55.845 g"/(6.022 * 10^(23)color(red)(cancel(color(black)("atoms Fe")))) = 9.2735 * 10^(-23)"g"#
#color(white)()# #color(purple)(2.)#
Now that you know the mass of one atom, you can use it to find the mass of #9.357 * 10^(30)# atoms
#9.357 * 10^(30)color(purple)(cancel(color(black)("atoms Fe"))) * (9.2735 * 10^(-23)"g")/(1color(purple)(cancel(color(black)("atom Fe")))) = color(green)(|bar(ul(color(white)(a/a)color(black)(8.677 * 10^8"g")color(white)(a/a)|)))#

The answer is rounded to four sig figs.

#color(white)()# ALTERNATIVE APPROACH

You can also find the answer by converting the number of atoms of iron to moles with the help of Avogadro's number, then by using iron's molar mass as a conversion factor.

You will have

#9.357 * 10^(30) color(red)(cancel(color(black)("atoms Fe"))) * "1 mole Fe"/(6.022 * 10^(23)color(red)(cancel(color(black)("atoms Fe")))) = 1.5538 * 10^7"moles Fe"#

This will once again be equivalent to

#1.5538 * 10^(7)color(red)(cancel(color(black)("moles Fe"))) * "55.845 g"/(1color(red)(cancel(color(black)("mole Fe")))) = color(green)(|bar(ul(color(white)(a/a)color(black)(8.677 * 10^8"g")color(white)(a/a)|)))#
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Answer 2

To find the mass of 9.357 * 10^30 atoms of iron:

  1. Find the molar mass of iron (Fe): 55.845 grams/mol.
  2. Calculate the number of moles of iron atoms: (9.357 \times 10^{30} , \text{atoms} \times \frac{1 , \text{mol}}{6.022 \times 10^{23} , \text{atoms}}).
  3. Multiply the number of moles by the molar mass of iron to find the mass: (9.357 \times 10^{30} , \text{atoms} \times \frac{1 , \text{mol}}{6.022 \times 10^{23} , \text{atoms}} \times 55.845 , \text{grams/mol}).

Performing the calculation:

[\text{Mass} = 9.357 \times 10^{30} \times \frac{1}{6.022 \times 10^{23}} \times 55.845 , \text{grams/mol} = 1.096 \times 10^8 , \text{grams}]

Therefore, the mass of 9.357 * 10^30 atoms of iron is approximately (1.096 \times 10^8) grams.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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