Question

what is the molar mass of $\ce{c_{5}h_{10}n_{2}o_{3}}$ show your work
% composition =
part / whole * 100
element | # | amu | total amu | % composition
c
h
n
o

mm

Explanation:

Step1: Determine the number of each element

For \( C_5H_{10}N_2O_3 \), the number of \( C \) atoms is 5, \( H \) is 10, \( N \) is 2, and \( O \) is 3.

Step2: Find the atomic mass (amu) of each element

• Atomic mass of \( C \): \( 12.01 \) amu
• Atomic mass of \( H \): \( 1.008 \) amu
• Atomic mass of \( N \): \( 14.01 \) amu
• Atomic mass of \( O \): \( 16.00 \) amu

Step3: Calculate total amu for each element

• For \( C \): \( 5 \times 12.01 = 60.05 \) amu
• For \( H \): \( 10 \times 1.008 = 10.08 \) amu
• For \( N \): \( 2 \times 14.01 = 28.02 \) amu
• For \( O \): \( 3 \times 16.00 = 48.00 \) amu

Step4: Sum the total amu to get molar mass

Molar Mass \( = 60.05 + 10.08 + 28.02 + 48.00 \)
\( = 146.15 \) g/mol (we can also use more precise values, but this is a common approximation)

Filling the table:

Element	#	Amu	Total amu	% composition
H	10	1.008	10.08	\( \frac{10.08}{146.15} \times 100 \approx 6.90\% \)
N	2	14.01	28.02	\( \frac{28.02}{146.15} \times 100 \approx 19.17\% \)
O	3	16.00	48.00	\( \frac{48.00}{146.15} \times 100 \approx 32.84\% \)
MM	-	-	146.15	-

Answer:

The molar mass of \( C_5H_{10}N_2O_3 \) is approximately \( 146.15 \) g/mol (or more precisely, using more accurate atomic masses, it can be calculated as well, but this is a standard result for this compound).