Question

pre - lab questions
answer these questions before arriving for lab. show your work fully and clearly.

1. draw the lewis structures of nitrate (no₃⁻) and phosphate (po₄³⁻). think carefully about how many electrons to use. then calculate the molar mass of each ion.
2. a sample has a nitrate concentration of 5.42 mg/l (5.42 ppm no₃⁻). calculate its molar concentration (in moles/l). (this primarily involves unit conversions, including using the molar mass of nitrate to convert mass to moles.)
3. a sample has a phosphate concentration of 0.00075 m. calculate the po₄³⁻ concentration in ppm.

Explanation:

Step1: Calculate molar mass of $NO_3^-$

N has an atomic mass of approximately 14 g/mol, O has an atomic mass of approximately 16 g/mol. In $NO_3^-$, molar mass $M_{NO_3^-}=14 + 3\times16=14 + 48 = 62$ g/mol.

Step2: Calculate molar mass of $PO_4^{3 -}$

P has an atomic mass of approximately 31 g/mol, O has an atomic mass of approximately 16 g/mol. In $PO_4^{3 -}$, molar mass $M_{PO_4^{3 -}}=31+4\times16=31 + 64 = 95$ g/mol.

Step3: Convert nitrate concentration to molar concentration (Question 2)

Given mass - concentration of $NO_3^-$ is $5.42$ mg/L = $5.42\times10^{- 3}$ g/L. Using the formula $n=\frac{m}{M}$, molar concentration $C=\frac{5.42\times10^{-3}\text{ g/L}}{62\text{ g/mol}}\approx8.74\times10^{-5}$ mol/L.

Step4: Convert phosphate concentration to ppm (Question 3)

Given molar - concentration of $PO_4^{3 -}$ is $0.00075$ M = $0.00075$ mol/L. Mass of $PO_4^{3 -}$ in 1 L is $m = n\times M=0.00075\text{ mol/L}\times1\text{ L}\times95\text{ g/mol}=0.07125$ g = $71.25$ mg. Since 1 ppm = 1 mg/L, the concentration in ppm is 71.25 ppm.

Answer:

1. Molar mass of $NO_3^-$: 62 g/mol; Molar mass of $PO_4^{3 -}$: 95 g/mol
2. Molar concentration of $NO_3^-$: $8.74\times10^{-5}$ mol/L
3. Concentration of $PO_4^{3 -}$ in ppm: 71.25 ppm