Question

A 50.0 mL sample of waste process water from a food processing plant weighing 50.1 g was placed in a properly prepared crucible that had a tare weight of 17.1234 g. After drying in a 103oC oven it weighed 19.9821 g. After the crucible was placed in a 550oC oven it weighted 18.7777 g. What are the TS and VS (in mg/L)?

Answer 1

Answer:

TS (total solid, mg/L) = 57,174 mg/L

VS (volatile solid, mg/L) = 24,088 mg/L

Explanation:

First step to solve this problem is to know data and questions:

Data:

Sample volume = 50 mL

Sample weight = 50.1 g

P1 (weight empty crucible) = 17.1234 g

P2 (weight after drying a 103º in oven) = 19.9821  g

P3 (weight after incinatrion 550º) = 18.7777 g

Questions:

TS = ?  

VS = ?

Formula:

We need to use formulas to calculate total solid (TS) and volatile solid (VS), these are:

$TS =\frac{(P2-P1)*\frac{1,000 mg}{1g} }{Sample volumen (L)}$

$VS = \frac{(P2-P3)*\frac{1,000 mg}{1g} }{Sample volume (L)}$

We have to transform mL to L so we will divide mL by 1,000 the sample volume:

$Sample Volumen (L) = 50 mL * \frac{1L}{1,000 mL} = 0.05 L$

In the formula the value of 1,000 results by the convertion factor to transform grams to miligrams (we have to multiplie by 1,000)

Now we need to replace data on previous formulas and we will get TS and VS expressed in mg/L:

$TS = \frac{(19.9821 g-17.1234g)*\frac{1,000 mg}{1g} }{0.05L} = 57,174mg/L$

$VS = \frac{(19.9821g-18.7777g)*\frac{1,000mg}{1g} }{0.05L}=24,088 mg/L$

We divide TS and VS formula by sample volume because the exercise is asking us to express the results in mg/L.