Final CHP111 Exam Fall2017 12/14/2017 ) Determine the empirical formulas of the

Question

Final CHP111 Exam Fall2017 12/14/2017 ) Determine the empirical formulas of the compounds with the following compositions by mass: (a) 55.3% K, 14.6% P, and 30.1% (b) 24.5% Na, 14.9% Si, and 60.6% F (c) 62.1% C, 5.2196 H, 12.196 N, and the remainder o (d)) Indicate the concentration of each ion present in the solution formed by mixing (a) 42.0 mL of 0.170 M NaOH and 37.6 mL of 0.400 M NaOH, (b) 44.0 mL of 0.100 M and Na2SO4 and 25.0 mL of 0.150 M KCl, (c) 3.60 g KCl in 75.0 mL of 0.250 M CaCl2 solution. Assume that the volumes are additive.

Explanation / Answer

(a)

1) Assume 100 g of the compound is present. This changes the percents to grams:

K 55.3 g
P 14.6 g
O 30.1 g

2) Convert the masses to moles:

K 55.3 g / 39.1 = 1.4143
P 14.6 g / 30.97 = 0.4714
O 30.1 g / 16 = 1.88125

3) Divide by the lowest, seeking the smallest whole-number ratio:

K 1.4143 / 0.4714 = 3
P 0.4714 / 0.4714 = 1
O 1.88125 / 0.4714 = 4

4) Write the empirical formula:

K3P1O4

(b)

1) Assume 100 g of the compound is present. This changes the percents to grams:

Na 24.5 g
Si 14.9 g
F 60.6 g

2) Convert the masses to moles:

Na 24.5 g / 23 = 1.07
Si 14.9 g / 28 = 0.532
F 60.6 g / 19 = 3.19

3) Divide by the lowest, seeking the smallest whole-number ratio:

Na 1.07 / 0.532 = 2
Si 0.532 / 0.532 = 1
F 3.19 / 0.532 = 6

4) Write the empirical formula:

Na2SiF6

(c)

1) Assume 100 g of the compound is present. This changes the percents to grams:

C 62.1 g
H 5.21 g
N 12.1 g
O 100 – (62.1 + 5.21 + 12.1) = 20.59 g

2) Convert the masses to moles:

C 62.1 g / 12 = 5.17
H 5.21 g / 1 = 5.21
N 12.1 g / 14 = 0.864
O 20.59 g / 16 =1.29

3) Divide by the lowest, seeking the smallest whole-number ratio:

C 5.17 / 0.864 = 6
H 5.21 / 0.864 = 6
N 0.864 / 0.864 = 1
O 1.29 / 0.864 = 1.5

4) Write the empirical formula:

C6H6N1O1.5

Converting to whole number, we get;

C12H12N2O3

(a)

42 mL of 0.170 M NaOH

Moles of NaOH = 0.170 M x 0.042 L = 0.00714 moles

37.6 mL of 0.400 M NaOH

Moles of NaOH = 0.400 M x 0.0376 L = 0.01504 moles

So, total moles of NaOH = 0.00714 moles + 0.01504 moles
                                        = 0.02218 moles

Total volume = 42 mL + 37.6 mL = 79.6 mL = 0.0796 L

So, [NaOH] = 0.02218 moles / 0.0796 L = 0.28 M

Since NaOH is a strong base and it will dissociate completely.

So,
[Na+] = [OH-] = 0.28 M

(b)

44 mL of 0.100 M Na2SO4

Moles of Na2SO4 = 0.100 M x 0.044 L = 0.0044 moles

1 mole of Na2SO4 produces 2 moles of Na+ and 1 mole of SO42-.

So,
moles of Na+ = 2 x 0.0044 moles = 0.0088 moles
moles of SO42- = 0.0044 moles

25 mL of 0.150 M KCl

Moles of KCl = 0.150 M x 0.025 L = 0.00375 moles

So,
moles of K+ = 0.00375 moles
moles of Cl- = 0.00375 moles

Total volume = 44 mL + 25 mL = 69 mL = 0.069 L

Hence

[Na+] = 0.0088 moles / 0.069 L = 0.1275 M
[SO42-] = 0.0044 moles / 0.069 L = 0.0638 M
[K+] = 0.00375 moles / 0.069 L = 0.0543 M
[Cl-] = 0.00375 moles / 0.069 L = 0.0543 M

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.