Calculate the number of C atoms, H atoms, and O atoms, in 1.50 grams of glucose (C6H1206).

Please explain and exact example of how to do this and what the correct answers are.

see below

To calculate the number of C atoms, H atoms, and O atoms in 1.50 grams of glucose (C6H12O6), we need to use the concept of moles and Avogadro's number.

Step 1: Find the molar mass of glucose (C6H12O6).
The molecular formula of glucose tells us that there are 6 carbon (C) atoms, 12 hydrogen (H) atoms, and 6 oxygen (O) atoms in one molecule of glucose.
The atomic masses are:
Carbon (C): 12.01 grams/mol
Hydrogen (H): 1.008 grams/mol
Oxygen (O): 16.00 grams/mol

To find the molar mass of glucose, we multiply the atomic mass of each element by its respective number of atoms in the molecule and sum them up:
Molar mass of glucose = (6 * 12.01 g/mol) + (12 * 1.008 g/mol) + (6 * 16.00 g/mol) = 180.18 g/mol

Step 2: Convert grams to moles.
To convert grams to moles, we divide the given mass of glucose (1.50 grams) by its molar mass (180.18 g/mol):
Number of moles = 1.50 g / 180.18 g/mol = 0.008325 moles (rounded to 4 decimal places)

Step 3: Use the mole ratio to find the number of atoms.
From the molecular formula of glucose (C6H12O6), we can see that there are 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms in one mole of glucose.
Multiply the number of moles (0.008325 moles) by the respective numbers of atoms:
Number of C atoms = 0.008325 moles * 6 C atoms/mole = 0.04995 moles (rounded to 5 decimal places)

Number of H atoms = 0.008325 moles * 12 H atoms/mole = 0.0999 moles (rounded to 4 decimal places)

Number of O atoms = 0.008325 moles * 6 O atoms/mole = 0.04995 moles (rounded to 5 decimal places)

Step 4: Convert the number of atoms to actual number of atoms.
To convert moles to the number of atoms, we use Avogadro's number, which is 6.022 x 10^23 atoms/mole.
Number of C atoms = 0.04995 moles * (6.022 x 10^23 atoms/mole) = 3.006 x 10^22 C atoms

Number of H atoms = 0.0999 moles * (6.022 x 10^23 atoms/mole) = 6.021 x 10^22 H atoms

Number of O atoms = 0.04995 moles * (6.022 x 10^23 atoms/mole) = 3.006 x 10^22 O atoms

So, in 1.50 grams of glucose (C6H12O6), there are approximately:
3.006 x 10^22 carbon (C) atoms,
6.021 x 10^22 hydrogen (H) atoms, and
3.006 x 10^22 oxygen (O) atoms.