Question

Aspartame, an artificial sweeterner, has the molecular formula C14H18N2O5. What is the mass in grams of one molecule?
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Answer 1

Answer:

Hi

Explanation:

This is a highly useful exercise in converting molar mass in grams to moles (which equates to 6.022x10^23 molecules according to Avogadro).

First we find the molar mass of Nutrasweet by looking up the molar mass of C (carbon), H (hydrogen), O (oxygen) and N (nitrogen) on a periodic table and multiply appropriately.

14*12.01+18*1.008+2*14.01+5*16=168.14+18.144+28.02+80=294.304g per mole aspartame

Now let's convert 1mg aspartame to moles:

1 mg * 1 g 1 mole

----------- = .001 g * ------------ = 3.398x10^-6 moles (yes these will be small amounts!)

1000 mg 294.304g

Now we can convert it to molecules:

3.398*10^-6 moles * 6.022*10^23 molecules

-------------------------- = 2.0462*10^18 molecules aspartame

1 mole

Now each molecule aspartame has 18 atoms of hydrogen so multiply that above number by 18:

3.6833*10^19 atoms hydrogen

Answer 2

Given:

The formula of aspartame is C₁₄H₁₈N₂O₅

To Find:

The mass of 1 molecule of Aspartame in grams

Solution:

• For solving this question we will apply the following law:

"1 mole of any substance is equal to its molar mass or equal to Avogadro's number (6.022 × 10²³) of atoms/molecules."           (1)

• We will follow these steps:

Step 1. Find the Molecular Mass of Aspartame.

• We know:

The molar mass of Carbon = 12 g

The molar mass of Hydrigen = 1 g

The molar mass of Nitrogen = 14 g

The molar mass of Oxygen = 16 g

• Thus, molar mass of Aspartame is:

C₁₄H₁₈N₂O₅ = 14(12) + 18(1) + 2(14) + 5(16)

                   = 294 g

Thus, molar mass of aspartame = 294 g.

Step 2. Find the mass of 1 molecule of Aspartame.

• From (1) we know that

                           294 g = 6.022 × 10²³ molecules of aspartame

∴Mass of 1 molecule is:

                                  M = $\frac{294}{6.022 .10^{23} }$

                              ⇒ M = 4.88 × 10⁻²² g

Thus, 1 molecule of Aspartame weighs 4.88 × 10⁻²² g.