How many moles of carbon atoms do you have if you have 47 g of carbon? Use your periodic table if necessary.
The Correct Answer and Explanation is:
Answer:
47 g of carbon contains approximately 3.91 moles of carbon atoms.
Calculation: Moles of carbon=Mass of carbon (g)Molar mass of carbon (g/mol)=4712.01≈3.91 mol\text{Moles of carbon} = \frac{\text{Mass of carbon (g)}}{\text{Molar mass of carbon (g/mol)}} = \frac{47}{12.01} \approx 3.91 \text{ mol}Moles of carbon=Molar mass of carbon (g/mol)Mass of carbon (g)=12.0147≈3.91 mol
Explanation
The mole is a fundamental concept in chemistry that provides a bridge between the atomic world and the macroscopic world. It allows chemists to count atoms by weighing them. One mole of any element contains 6.022×10236.022 \times 10^{23}6.022×1023 entities (Avogadro’s number), whether atoms, molecules, or other particles.
To determine how many moles of a substance are present in a given mass, the mass must be divided by the molar mass of the substance. The molar mass is the mass of one mole of an element or compound, expressed in grams per mole (g/mol), and can be found on the periodic table. For carbon, the average atomic mass is approximately 12.01 g/mol. This value takes into account the natural abundance of the carbon isotopes, primarily carbon-12 and carbon-13.
In this case, we are given a mass of carbon equal to 47 grams. Dividing this mass by the molar mass of carbon gives the number of moles: 47 g12.01 g/mol=3.91 mol\frac{47 \text{ g}}{12.01 \text{ g/mol}} = 3.91 \text{ mol}12.01 g/mol47 g=3.91 mol
This means that 47 grams of carbon is equivalent to 3.91 moles of carbon atoms. Since one mole of carbon atoms contains 6.022×10236.022 \times 10^{23}6.022×1023 atoms, multiplying the number of moles by Avogadro’s number would give the total number of atoms, although that is not required for this calculation.
Understanding how to convert between mass and moles is essential in chemical calculations, including determining the amounts of reactants and products in chemical reactions, making solutions, and studying stoichiometry.
