Final answer:
To find the hypothetical molar mass of 'M&Ms' conceptually treated as an element, one would calculate the mass of one 'M&M' and multiply by Avogadro's number. This results in an impractically large number, illustrating why this approach is purely theoretical and not applicable in actual chemistry.
Step-by-step explanation:
To calculate the atomic mass of 'M&Ms' if they were to be treated conceptually as an element and each 'M&M' represented an atom, you would take the total mass of the bag and divide it by the number of 'M&Ms' to get the mass of one 'M&M'. Then, you would convert this mass to grams per mole (g/mol) using Avogadro's number (6.022 × 1023), which is the number of particles in one mole of a substance.
First, convert the mass of the bag from ounces to grams, knowing that 1.75 oz is approximately 49.61 g (since 1 oz = 28.35 g). If there are about 55 'M&Ms' in a bag, then the mass of one 'M&M' would be 49.61 g ÷ 55, which is about 0.902 g per 'M&M'.
However, since the molar mass is the mass of 1 mole, you would then consider how many 'M&Ms' would be in one mole. This is a theoretical scenario, as 'M&Ms' are not actual atoms, but for the sake of the question, if one mole of 'M&Ms' roughly contained 6.022 × 1023 'M&Ms', then the molar mass of an 'M&M' element would be 0.902 g × 6.022 × 1023, which is an enormously large number and not practical in real chemistry terms.
In practical chemistry, the atomic mass and molar mass are numerically equivalent but differ by scale - the former is measured in atomic mass units (amu) and the latter in grams/mole. For example, the molar mass of carbon is 12.01 g/mol, which indicates that 1 mole of carbon atoms weighs 12.01 grams.