A volume of 13.96 mL of 0.1060 M NaOH solution was used to titrate a 0.618 g sam

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. ?

Explanation / Answer

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

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