Question

How many g of caco3 are present in a sample if there are 4.52 x 1024 atoms of carbon in that sample?

Answer 1

Answer:
mass of CaCO3 = 750.5 grams

Explanation:
One mole of any substance contains Avogadro's number of atoms.
This mean that "x" moles of a substance will contain x * Avogadro's number atoms.
We are given that the sample contains 4.52 * 10^24 carbon atoms. 
Therefore, we can get the number of moles of carbon in the sample as follows:
number of atoms = number of moles * Avogadro's number
Therefore:
number of moles = number of atoms / Avogadro's number
number of moles = (4.52 * 10^24) / (6.022 * 10^23) = 7.505 moles

Now, from the given compound, we can note that one mole of CaCO3 contains one mole of carbon.
This means that 7.505 moles of carbon will be present in 7.505 moles of CaCO3

Final step is to get the mass of 7.505 moles of CaCO3 as follows:
From the periodic table:
molar mass of calcium = 40 grams
molar mass of carbon = 12 grams
molar mass of oxygen = 16 grams
Therefore,
molar mass of CaCO3 = 40 + 12 + 3(16) = 100 grams/mole
This means that:
mass of 7.505 moles of CaCO3 = 7.505 * 100 = 750.5 grams

Hope this helps :)

Answer 2

Answer:

750.8 grams of CaCO₃ are present in a sample if there are 4.52*10²⁴ atoms of carbon in that sample.

Explanation:

Avogadro's Number or Avogadro's Constant is called the number of particles that make up a substance (usually atoms or molecules) and that can be found in the amount of one mole of said substance. Its value is 6.023 * 10²³ particles per mole. The Avogadro number applies to any substance.

So, the following rule of three applies: if 6.02 * 10²³ carbon atoms are contained in 1 mole of said compound, 4.52*10²⁴ atoms in how many moles will they be contained?

$moles of carbon=\frac{4.52*10^{24}atoms*1mole }{6.02*10^{23} atoms}$

moles of carbon=7.508

For the given compound, you can see that one mole of CaCO₃ contains one mole of carbon.  This means that 7,508 moles of carbon will be present in 7,508 moles of CaCO₃.

To know the mass you must know the molar mass of the compound. You know what:

• Ca: 40 g/mole
• C: 12 g/mole
• O: 16 g/mole

So: CaCO₃= 40 g/mole + 12 g/mole + 3*16 g/mole= 100 g/mole

So, you apply a rule of three as follows: if one mole of the compound contains 100 g, 7,508 moles of CaCO₃, how much mass will it contain?

$mass=\frac{7.508moles*100 g}{1 mole}$

mass= 750.8 grams

750.8 grams of CaCO₃ are present in a sample if there are 4.52*10²⁴ atoms of carbon in that sample.