CHEMISTRY 311 - ASSIGNMENT 1 1 2018 solutions.pdf · CHEM 311 Assign 1 2018 - Solutions Now we are...

CHEM 311 Assign 1 2018 - Solutions

CHEMISTRY 311 - ASSIGNMENT 1

Hand-in your answers in a neat organized format to the appropriate number significant figures, showing chemical equations and calculations where appropriate.

Due date: Tuesday, September 25th

1. Calculate the mass of primary standards (NaNO3 and KH2PO4

.3H2O) required to prepare 250.0 mL of a combined standard solution containing 100.0 µg/L of NO3

--N and 10.00 µg/L of PO43--P. Are there any practical

challenges that limit the precision of the reported concentrations of your combined standard solution? If so, how might you suggest modifying your procedure. Strategy: A primary standard is a high purity chemical compound of known composition used in calibration either directly or by preparing standard solutions. This question involves calculating the mass of two different salts one containing nitrogen and the other phosphorus that will be dissolved and added to the solution to prepare a combined standard. Since the desired concentrations are reported in terms of the mass of N and P (respectively), we will need to use the mass ratio of nitrogen in NaNO3 and that of phosphorus in KH2PO4.3 H2O. Note that the water molecules of hydration in the phosphorus salt are included in the molar mass calculation.

Solution:

𝑚𝑎𝑠𝑠 𝑁𝑎𝑁𝑂3 =100.0 𝑢𝑔 𝑁

𝐿 𝑥84.9946 𝑔 𝑁𝑎𝑁𝑂314.0067 𝑔 𝑁 𝑥 0.2500𝐿

= 151.703 𝑢𝑔 𝑁𝑎𝑁𝑂3 𝑜𝑟 0.151703 𝑚𝑔 𝑁𝑎𝑁𝑂3

𝑚𝑎𝑠𝑠 𝐾𝐻2𝑃𝑂4.3𝐻2𝑂 =10.00 𝑢𝑔 𝑃

𝐿 𝑥 190.1314 𝑔 𝐾𝐻2𝑃𝑂4.3𝐻2𝑂

30.9737 𝑔 𝑃 𝑥 0.2500𝐿

= 15.3462 𝑢𝑔 𝐾𝐻2𝑃𝑂4.3𝐻2𝑂 𝑜𝑟 0.0153462 𝑚𝑔 𝐾𝐻2𝑃𝑂4.3𝐻2𝑂

Answer: 0.1517 mg NaNO3 0.01534 mg KH2PO4.3H2O

Given that a typical analytical mass balance has a reading error of +/- 0.1 mg, these masses are too small to weigh out an accurate or precise amount of either salt. One strategy here would be to prepare a 1L stock solution containing 151.7 mg of NaNO3 and 15.3 mg of KH2PO4.3H2O. This will be 1000 times more concentrated than needed, so if one pipetted 1.00 mL into a fresh 1 L volumetric flask and dilute to the mark, we would have the desired concentration. In preparing analytical standards, we typically aim to weigh out ~ 100 mg quantities of primary standards to minimize uncertainties resulting from reading errors.


2. Sodium sulfite (Na2SO3) is sometimes used to remove excess chlorine from disinfected water supplies, especially if they are going to be used for fish rearing. a) Write a balanced redox reaction in which the final products are sulfate and chloride ion. b) Calculate the mass of sodium sulfite required to react with the residual chlorine in 26 m3 of supply water containing 1.2 mg/L Cl2? Note that sodium sulfite also reacts with dissolved oxygen, so adding too much to a fish tank can have dire consequences, so it is important to get the math and chemistry right on this one.

Strategy: This question requires writing a balanced redox reaction between sulfite ion and chlorine to yield sulfate and chloride ions. Once this has been accomplished, we will use the mole ratio between chlorine and sulfite ion to determine the number of moles of Na2SO3 required to react with all of the Cl2 in 26 m3 containing 1.2 mg/L.

Solution: Balancing redox reactions starts by identifying the chemical species undergoing a change of oxidation state. In this case, the sulfur atom in sulfite to being oxidized from +4 to +6 in sulfate. Similarly, the chlorine atoms are being reduced from 0 to -1. We separate the two half reactions and balance each as follows;

i) Add required number of electrons

SO32- à SO4

2- + 2 e-

ii) Balance the number of atoms that have changed oxidation state

Done

iii) Balance oxygen atoms by adding H2O molecules

SO32- + H2O à SO4

2- + 2 e-

iv) Balance the number of hydrogen atoms by adding H+

SO32- + H2O à SO4

2- + 2e- + 2 H+

v) Check for charge balance

Following the same steps for the other half reaction yields,

Cl2 + 2 e- à 2 Cl-

Now recombine the two half reactions ensuring that the number of electrons transferred is the same.

SO32- + Cl2 + H2O à SO4

2- + 2 Cl- + 2 H+


Now we are in a position to solve part b) and determine the mass of Na2SO3 required. Working backwards, we can say that the mass of sodium sulfite will be equal to the number of mols of Na2SO3 required times the molar mass of Na2SO3, which is 126.04 g/mol.

𝑚𝑎𝑠𝑠 𝑁𝑎2𝑆𝑂3 = 𝑛!"!!"! 𝑥 𝑀𝑀!"!!"! = 𝑛!"!!"! 𝑥 126.04 𝑔/𝑚𝑜𝑙

We can then determine the number of moles of Na2SO3 required using the balanced redox reaction from part a).

𝑛!"!!"! = 𝑛!"!𝑥 1 𝑚𝑜𝑙 𝑆𝑂32−1 𝑚𝑜𝑙 𝐶𝑙2 𝑥

1 𝑚𝑜𝑙 𝑁𝑎2𝑆𝑂31 𝑚𝑜𝑙 𝐶𝑙2

The number of mols of Cl2 can be calculated using the concentration of 1.2 mg Cl2/L and the volume of water being treated (26 m3).

𝑛!"! =1.2 𝑚𝑔 𝐶𝑙2

𝐿 𝑥 1 𝑚𝑜𝑙70.91𝑔 𝑥

1 𝑔10! 𝑚𝑔 𝑥

10! 𝐿𝑚! 𝑥 26 𝑚3 = 0.4399 𝑚𝑜𝑙𝑠 𝐶𝑙2

Therefore, the mass of Na2SO3(s) required is given by;

0.4399 mols Na2SO3 x 126.04 g Na2SO3/mol = 55.457 g

Answer: 55 g Na2SO3


3. A mineral is analyzed for it’s calcium content as follows. A 1.032 g finely ground sample was dissolved in 25 mL of hot 4 M HCl and diluted with 175 mL of H2O. This solution was heated to a boil and 50 mL of a hot solution containing 2.0 g of (NH4)2C2O4 was slowly added to precipitate CaC2O4(s) as the monohydrate. The solution was neutralized by adding 6 M NH3 and slowly cooled. After decanting the supernatant solution, the solid was transferred to a filter crucible and washed with several portions of cold 0.1% (NH4)2C2O4 . The crucible was dried at 105oC for 1 hr and then heated at 500oC in a furnace for an additional 2 hr. The mass of the precipitate was recorded to be 0.154 g.

105oC 500oC

Ca2+ + C2O42- à CaC2O4 . H2O(s) à CaCO3(s)

a) Determine the concentration of calcium in the mineral sample in mg/g. b) Provide three desired properties of a precipitate in gravimetric analysis.

Strategy: This question involves gravimetric analysis in which the mass of the final precipitate is used to determine the amount of calcium present in the original solid sample. There are a series of chemical transformations to get the original sample to dissolve, and then react with oxalate ion (C2O4

2-) which is isolated and then converted at high temperature to calcium carbonate.

Solution:To determine the mass of Ca in the sample in mg/g, we need the mass calcium in the final precipitate and the original sample.

𝐶𝑜𝑛𝑐 𝐂𝐚 𝑚𝑔𝑔 =

𝑚𝑎𝑠𝑠 𝐶𝑎 𝑖𝑛 𝑝𝑟𝑒𝑐𝑖𝑝𝑖𝑡𝑎𝑡𝑒 (𝑚𝑔) 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑎𝑚𝑝𝑙𝑒 (𝑔)

Since the mass of sample was given as 1.032 g, the only mass we need to calculate in the mass of calcium in the 0.154 g of CaCO3(s). We might use the fact that CaCO3 is 40.0% Ca by mass (which we determine using the relative molar mass of Ca to CaCO3).

In other words,

𝑚𝑎𝑠𝑠 𝐶𝑎 = 0.154 𝑔 𝑥 !".!""# ! !"/!"#!"".!"#$ ! !"!#!/!"#

𝑥 !"! !"! !

= 61.6664 mg

Conc Ca = 61.6664 mg/1.032 g = 59.754 mg/g

Answer:

59.8 mg Ca/g

insoluble, easily filterable and known composition (cf, textbook p.630)


4. The alkalinity of a freshwater sample is a measure of its acid neutralizing ability and is primarily due to the presence of hydrogen carbonate ion (HCO3

-). The Total Alkalinity is determined by titrating a water sample with a standardized acid to a methyl orange endpoint (pH ~ 4.5). What volume of 0.0280 N H2SO4 titrant is required to titrate a 25.0 mL water sample with a known alkalinity of 120. ppm of CaCO3? Does your result suggest a problem with the methodology? If so, how could you modify the procedure to reduce the relative uncertainty in your analytical result?

Strategy: This question involves the volumetric determination of Total Alkalinity. We need to calculate the volume of 0.0280 N H2SO4 titrant required to neutralize a 25.0 mL sample with a total alkalinity of 120 ppm CaCO3 to a pH of 4.5. Since the concentration of the titrant is given as Normality, we can immediately say that the #equivalents of titrant = #equivalents of analyte. Hence, we can write,

𝑉!"!#$%! =𝑁!"!#$%& 𝑥 𝑉!"!#$%&

𝑁!"!#$%!

Solution:

𝑉!"!#!"# =𝑁!"!#$%& 𝑥 25.0 𝑥 10!!𝐿

0.0280 𝑒𝑞/𝐿

Converting the analyte concentration from ppm CaCO3 to Normality using an equivalent weight of 50,000 mg/eq as follows;

𝑁!"!#$%& = 120.𝑚𝑔𝐶𝑎𝐶𝑂3𝐿 𝑥

1 𝑒𝑞50,000 𝑚𝑔 = 2.40 𝑥 10!! 𝑒𝑞/𝐿

Therefore, the volume required to reach the end-point is 2.1428 x 10-3 L or 2.143 mL

Answer:

2.14 mL of titrant

Given that each volume reading in a class A 50 mL buret is +/-0.02 mL, the propagated reading error in titrant volume is +/-0.03 mL. Hence the relative uncertainty in the total alkalinity of this sample will be at least (0.03/2.14) x 100% = 1.4%.

To reduce the relative uncertainty in the method, we need to have a larger titrant volume. In volumetric analysis, we aim to have titrant volumes greater than 10 mL. In this case, we can achieve this by titrating a larger sample volume and/or using a more dilute titrant.

The equivalent weight of calcium carbonate is equal to the mass of CaCO3 that will neutralize (react with) one mol of H+. Since, CaCO3 + 2 H+ à H2CO3 + Ca2+

K (for CaCO3 as a base) = 2 mols H+/1 mol CaCO3 = 2 eq/mol. Therefore, the EW = MW/2 = 50.0 g/eq


5. The analysis of a Reference Material for lead in blood samples with a certified concentration of 12.6 ng/mL was analyzed by graphite furnace atomic absorption (GFAA) spectroscopy. The results of six replicate analysis are listed below. Check for potential outliers using the Grubb’s test at the 95% confidence level. Incorporating the removal of outliers (if any), quantify the precision and accuracy of the GFAA method to analyze lead in blood samples.

12.9 11.2 10.6 12.1 13.0 11.5 ng/mL

Strategy: This question involves quantitative measures of both precision and accuracy. Precision is measure of ‘reproducibility’ and is best assessed by calculating the sample standard deviation and converting this to a relative standard deviation and a confidence interval (typically at the 95% confidence level). Precision is limited by random indeterminant errors. Accuracy is measured by comparing the experimental mean to the accepted value and calculating a % difference (also known as the % bias). Differences between the mean experimental value and an accepted one are due to determinant (systematic) errors.

Solution:

The Grubb’s test uses the ratio of [(suspect value – mean)/sample standard deviation].

We can test either the maximum (13.0) or minimum (10.6) values as potential outliers by computing a corresponding Gcalc and then compare this to a tabulated critical value at the same confidence level and the appropriate number of observations (see further Appendix 2 Lab Manual or Chap 3 of textbook).

Gtab = 1.822 for N=6

The sample standard deviation is given by;

€

s =(xi − x)

2∑N −1

For the original given data set;

N=6,

€

x = 11.883 ng/mL, s = 0.95795 ng/mL

Therefore,

𝐺!"#! =13.0− 11.880.9579 = 1.166

𝐺!"#! =10.6− 11.880.9579 = 1.366

Since both extrema values are less than Gtab, all values are retained.


Precision Hence, the relative standard deviation (RSD) = (0.95795/11.883) x 100% = 8.1% (note rounding uncertainty to two significant figures is customary)

The confidence interval is given by;

𝐶𝐼 = ! !!

= !.!"# !.!"#!"!

= 1.005 ng/mL

where t = 2.x571 at N-1 = 5 degrees of freedom and the 95% confidence level.

Consequently, we can be 95% confident that our experimental value lies between is 10.88 and 12.88 +/- ng/mL.

Accuracy The % difference (bias) between our experimental mean and the accepted value is given by;

% 𝑏𝑖𝑎𝑠 = 11.883− 12.6

12.6 𝑥 100% = −5.7 %

As we can see, the precision of the analysis of this sample about 8% as reported by the RSD. The accuracy of the determination is about negative 6% (meaning that the experimental value is less than the true (or accepted) value. It should be noted however, that the lack of precision limits our ability to be confident that this bias is real.


6. In the standardization of hydrochloric acid, high purity sodium carbonate is dried and weighed into a flask and dissolved in water before being titrated with HCl.

Mass of Na2CO3 (±0.1 mg)

Volume of HCl delivered (±0.03mL)

126.0 23.77 105.4 19.51 122.8 22.85 112.5 20.06

a) What is the balanced chemical reaction that occurs. b) Write a general mathematical expression for the millimolar concentration of HCl in terms of the mass of Na2CO3 (mg) and the volume of HCl titrant (mL). c) What is the absolute and relative uncertainty in the calculated concentration based on the propagation of the reported reading errors for titration of 105.4 mg of Na2CO3? d) Calculate the mean concentration, standard deviation and relative standard deviation based on the data from the four trials.

Strategy: This question involves several steps. Part b) has us derive a general expression that can be used to carry out the calculation of the HCl concentration from mass of Na2CO3 and volume of HCl titrant as reported in the table above.

Solution:

a) Balance chemical reaction

Na2CO3(aq) + 2 HCl(aq) à H2CO3(aq) + 2 NaCl(aq)

Note that the Na+ and Cl- are spectator ions and not involved in any net chemistry. Therefore, we could also write the net reaction as;

CO32-(aq) + 2 H+(aq) à H2CO3(aq)

b) General expression in mM

𝐻𝐶𝑙 𝑚𝑀

= 𝑚𝑎𝑠𝑠 𝑁𝑎2𝐶𝑂3 𝑚𝑔 𝑥 1 𝑔

10! 𝑚𝑔 𝑥 1 𝑚𝑜𝑙

105.9884 𝑔 𝑥 2 𝑚𝑜𝑙 𝐻𝐶𝑙

1 𝑚𝑜𝑙 𝑁𝑎2𝐶𝑂3 𝑥1

𝑉𝑜𝑙𝑢𝑚𝑒 𝐻𝐶𝑙 (𝑚𝐿) 𝑥 10! 𝑚𝐿1 𝐿 𝑥

10!𝑚𝑀1 𝑀

Consolidating all constants yields;

𝐻𝐶𝑙 𝑚𝑀 = 18.8699 (𝑚𝑎𝑠𝑠 𝑁𝑎2𝐶𝑂3 𝑖𝑛 𝑚𝑔)

(𝑣𝑜𝑙𝑢𝑚𝑒 𝐻𝐶𝑙 𝑖𝑛 𝑚𝐿)


c) Propagation of reading error The rules for the propagation of errors depends on the mathematical functions involved in the calculation. In the case of multiplication and division, it is the relative errors that are additive.

[HCl] (mM) = 18.8699 (105.4 +/- 0.1 mg)/(19.51 +/- 0.03) = 101.942 +/- propagated error (mM)

𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑒𝑟𝑟𝑜𝑟 = 0.1105.4

!

+ 0.0319.51

!

= 9.00 𝑥 10!! + (2.36 𝑥 10!!) = 1.81 𝑥 10!!

or 0.181 %

Converting this back to an absolute error, involves multiplication of this relative error by the absolute value of the concentration of HCl (in other words 0.181% of 101.942 mM).

Thus, the absolute error in the concentration of HCl based on the propagation of reading errors alone from the second trial is 0.18 mM.

101.94 +/- 0.18 mM

d) Overall

Applying the general expression in part b) to the four entries in the table above yields

100.026 mM

101.942 mM

101.162 mM

105.826 mM

If we retain all values, we get the following;

Mean = 102.239 mM

Std dev = 2.5 mM

Relative std dev = 2.5%

Note that the uncertainty derived from the propagation of reading errors for individual trials is considerably less than the uncertainty based on actual measurement replicates. This is generally the case due to additional sources of variance in the replication of the measurement (including the operator). In general, when multiple measurements have been performed, it the always the larger uncertainty estimate that is reported. If a measurement has only been done once, then the propagated reading error is the only estimate of uncertainty available.


7. A field portable volumetric analysis method for Dissolved Oxygen using a Hach Digital Titrator™ is based on the Winkler titration, whereupon the dissolved oxygen is converted to an equivalent amount of iodine which is then titrated with sodium thiosulfate. The volume of titrant added is recorded with a digital counter (corresponding to the delivery of 1.25 µL) and can be converted to an analyte concentration with an appropriate ‘digit multiplier’. a) Calculate the concentration of the thiosulfate titrant required such that the ‘digit multiplier’ for the titration of a 60. mL sample is 0.10 ppm O2 per digit. b) What mass of Na2S2O3.5H2O would be required to prepare 100. mL of this titrant? Strategy: This question involves the volumetric analysis of dissolved oxygen with a portable titrator. If we work in Normality, we can directly relate the number of equivalents of the analyte to that of the titrant.

NO2 x VO2 = NS2O32- x VS2O32-

Since one digit increment corresponds to 1.25 x 10-6 L of titrant and 0.10 ppm O2, we can convert the oxygen concentration to Normality.

Solution: We can calculate the Normality of thiosulfate as follows

𝑁!!!!"! =!!! !!!!!!!!"!

= (!.!" ! !"!!!"! )(!".! !"!!!)

!.!" ! !"!! ! = 0.6000 eq/L

where 0.10 ppm O2 was converted to 1.25 x 10-5 eq/L using an equivalent weight for O2 = 8,000 mg/eq.

The equivalent weight of O2 is the mass required to accept one mol of electrons as O2 à H2O. Since this is a 4 e- process, the K factor for O2 in this case is 4 mol e-/mol O2 = 4 eq/mol and the EW = MW/4 = 8.00 g/eq.

Mass of Na2S2O3.5H2O to prepare 100. mL of 0.6000 N solution is given by;

𝑚𝑎𝑠𝑠 = 0.6000 𝑒𝑞

𝐿 𝑥 248.1841 𝑔

𝑒𝑞 𝑥 0.100 𝐿 = 14.8910 𝑔 𝑁𝑎2𝑆2𝑂3.5𝐻2𝑂

where the K factor for thiosulfate (S2O32-) reacting to produce tetrathionate ion (S4O6

2-) is 1 eq/mol.

Since, S2O32- à S4O6

2- involves the transfer of one mol e- per mol of S2O32-

Answer:

NS2O3 = 0.600 eq/L

Mass = 14.9 g


8. The concentration of cyanide (CN-) in a copper electroplating bath can be determined by a complexation titration with Ag+, forming the soluble Ag(CN)2

- complex. In a typical analysis a 5.00 mL sample from an electroplating bath is transferred to a 250 mL Erlenmeyer flask and treated with 100 mL of H2O, 5 mL of 20% w/v NaOH and 5 mL of 10%w/v KI. A sample was titrated with 0.2012 M AgNO3, requiring 17.36 mL to reach the end-point as signaled by the appearance of a yellow precipitate of AgI. a) Suggest the appropriate lab glassware to be employed for transferring each of the sample, water, sodium hydroxide and potassium iodide solutions, respectively. b) What is the molarity of a 10% w/v KI solution and why is it added? c) Report the concentration of cyanide in the electroplating bath as parts per million of CN-. Strategy: This question involves a volumetric analysis of cyanide ion using a standard solution of silver ions as the titrant. The end point is reached at the first observation of a yellow precipitate which occurs after the analyte has been consumed and the remaining Ag+ combines with the excess I- to form AgI(s).

Ag+ + 2 CN- à Ag(CN)2-

Solution:

a) The 5.00 mL of sample will be measured precisely using a class A pipet. The 100 mL water is not crucial to the quantitative analysis and can be measured imprecisely with a beaker or the Erlenmeyer flask itself. The NaOH and KI solutions are added in vast excess and their volumes can be measured imprecisely with a beaker or graduated cylinder. Finally, the titration volume will be precisely measured using a class A 50mL buret.

b) The KI solution is reported to be 10% (w/v). This means that there is 10 g of NaOH(s) dissolved into 100 mL of solution. We can convert this to molarity using the molar mass of NaOH as follows;

𝐾𝐼 = !" ! !"#$!"" !" !"#$

𝑥 ! !"#!"".!! ! !"

𝑥 !"! !"! !

= 0.602 mol/L

added to combine with excess Ag+ to form yellow precipitate as end-point indicator

c) Concentration of CN- (ppm) = [CN-] x 26.0174 g/mol x (103 mg/g) x (1 L/1.00 kg)

𝐶𝑁! = 𝑛!"! 𝑥 2 𝑚𝑜𝑙 𝐶𝑁!

1 𝑚𝑜𝑙 𝐴𝑔! 𝑥 1

5.00 𝑚𝐿 𝑥 10! 𝑚𝐿1 𝐿

= 𝑀!"! 𝑥 𝑉!"! 𝑥 2 𝑚𝑜𝑙 𝐶𝑁!


5.00 𝑚𝐿 𝑥 10! 𝑚𝐿1 𝐿

= 0.2012𝑚𝑜𝑙𝐿 𝑥 17.36 𝑚𝐿 𝑥

1 𝐿10! 𝑚𝐿 𝑥

2 𝑚𝑜𝑙 𝐶𝑁!


5.00 𝑚𝐿 𝑥 10! 𝑚𝐿1 𝐿 = 1.39954 𝑚𝑜𝑙/𝐿

Converting this to ppm CN- yields,

36,412.57 ppm CN-

Answer:

36,400 ppm CN-


BONUS. The Kjeldahl procedure was used to analyze a 5.00 mL of a solution containing 390 mg of a food sample. The liberated NH3 was collected in 50.00 mL of 0.0336 M HCl, and the remaining acid was back-titrated with 12.34 mL of 0.1046 M NaOH for complete titration. a) What chemistry is involved in the Kjeldahl procedure? b) What is the mass percent of nitrogen in the sample? Strategy: This question involves a volumetric back titration of the ammonia produced in the digestion step. The amount of ammonia produced is equal to the difference between the initial number of mols of HCl and the remaining number of mols of HCl from the titration with NaOH. Solution:

a) The chemistry involved in the Kjeldahl procedure involves heating a sample to around 350 oC in sulfuric acid solution in the presence of potassium sulfate. After adding enough sodium hydroxide to neutralize the acid and convert ammonium ion to ammonia, the NH3 is distilled into a solution with a known excess amount of H+. The remaining H+ is back-titrated to determine the amount of ammonia released in the digestion.

Sample + H2SO4 à (NH4)2SO4(aq) + CO2(g) + SO2(g) + H2O

(NH4)2SO4(aq) + NaOH à Na2SO4(aq) + 2 H2O + 2 NH3(g)

b) The mass percent of nitrogen in the sample = (mass of N/mass of sample) x 100%

Mass of N = mols N x 14.0067 g/mol

𝑛!"! = 𝑛!! (reacted) = nH+ (initially) – nH+ (remaining)

nH+ (initially) = MHCl x VHCl = 0.0336 mol/L x 50.00 x 10-3 L = 1.68 x 10-3 mols

nH+ (remaining) = nNaOH added = MNaOH x VNaOH = 0.1046 mol/L x 12.34 x 10-3 L = 1.29076 x 10-3 mols

Therefore,

𝑛!! (reacted) = (1.680 – 1.29076) x 10-3 mols = 0.38924 x10-3 mols = nNH3

Mass N = 3.8924 x 10-4 mols x (1 mol N/1 mol NH3) x 14.0067 g N/mol = 5.4519 x 10-3 g = 5.4519 mg

Mass percent N in sample = (5.4519 / 390) x 100% = 1.3979%

Answer:

1.4% N in sample

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