Question

A volume of 13.96 mL of 0.1060 M NaOH solution was used to titrate a 0.618 g sample of unknown containing HC₇H₅O₃.

What is the molecular mass of HC₇H₅O₃? (report answers to 4 or 5 significant figures) 1.3812×10² g/mol

What is the percent by mass of HC₇H₅O₃ in the unknown?

In this problem what mass of sample in grams would be needed to deliver about 23.40 mL in the next trial?

In the second trial above, exactly 1.032 g was transferred into a flask to be titrated. If the initial buret reading is 0.10 mL, predict what the final buret reading be. ?

Answer 1

STEP 1: CALCULATION OF NUMBER OF MOLES OF NAOH IN 13.96 mL of 0.1060 M NaOH SOLUTION

1000 mL of 1 M NaOH contains 1 mol NaOH

13.96 mL of 0.1060 M NaOH contains 1 x 13.96 x 0.1060 / 1000 x 1 = 0.00148 mol NaOH

STEP 2: CALCULATION OF NUMBER OF MOLES OF HC₇H₅O₃ IN 0.618 g sample of unknown containing HC₇H₅O₃.

HC₇H₅O₃ is monoprotic/ monobasic acid (Note that one H atom is wrtten separately)

Hence 1 mol NaOH will react with 1 mol HC₇H₅O₃

Therefore 0.00148 mol NaOH will react with 0.00148 mol HC₇H₅O₃

Hence NUMBER OF MOLES OF HC₇H₅O₃ IN 0.618 g sample of unknown containing HC₇H₅O₃ = 0.00148

STEP 3: CALCULATION OF MOLECULAR MASS OF HC₇H₅O₃

HC₇H₅O₃ = C₇H₆O₃

Molecular mass of HC₇H₅O₃ = C₇H₆O₃ = 7 x12 .010 + 6 x1.008 + 3 x 16.000 = 138.118

1 mole o f a substance is its molecular mass expressed in gram

Therefore 1 mole of HC₇H₅O3 = 138.118 g

or Molecular mass of HC₇H₅O3 = 138.118 g/mol = 1.3812 x 10² g/mol

STEP4: CALCULATION OF ACTUAL MASS OF HC₇H₅O₃ IN 0.618 g sample of unknown containing HC₇H₅O₃.

NUMBER OF MOLES OF HC₇H₅O₃ IN 0.618 g sample of unknown containing HC₇H₅O₃ = 0.00148

1 mole of HC₇H₅O3 = 138.118 g

Therefore 0.00148 mole of HC₇H₅O3 = 0.00148   x 138.118 g = 0.2044 g

STEP4: CALCULATION OF the percent by mass of HC₇H₅O₃ in the unknown

0.618 g sample of unknown contains 0.2044 g of HC₇H₅O₃.

Therefore100 g sample of unknown contains 0.2044 g x 100 g / 0.618 g of HC₇H₅O₃ = 33.07 g of HC₇H₅O₃

The percent by mass of HC₇H₅O₃ in the unknown = 33.074 %

STEP 5 CALCULATION OF mass of sample in grams that would be needed to deliver about 23.40 mL in the next trial

13.96 mL of 0.1060 M NaOH solution requires 0.618 g sample

Therefore 23.40 mL of 0.1060 M NaOH solution requires 0.618 g x 23.40 / 13.96 = 1.036 g sample

STEP 5 PREDICTION OF FINAL BURETTE READING

0.618 g sample requires 13.96 mL of 0.1060 M NaOH solution

Therefore 1.032 g sample requires 13.96 mL x 1.032 / 0.618 = 23.31 mL of 0.1060 M NaOH solution

Initial reading is 0.10 mL

Final reading will be 0.10 mL + 23.31 mL = 23.41 mL