Do Question Number 6 SHoW YOUR WORK VERY, VERY CLEARL.Y 1. A mixture of 3.5 g of

Question

Do Question Number 6

SHoW YOUR WORK VERY, VERY CLEARL.Y 1. A mixture of 3.5 g of hydrogen and 260 g of oxygen is made to react to form water a) Which reactant is limiting? b) Which reactant will be left over? ) How many reactant will be left over? d) How many grams of water will be grams of the produced? A large neon sign is to be filled with a mixture of gases, including 6348 g of neon. What number of mole is this? 2. A metal, M, forms an oxide having the simplest formula M,Os. This oxide contains 52.9 % of metal by mass. a) Calculate the atomic weight of the metal; b) Identify the metal. 3, 4. From a 45.0 g sample of an iron ore containing Fe,O 1.5 g Fe is obtained by the following reaction: Fe,04 + 2C 3Fe + 2CO2. What is the percent if Fe,0? 5. Balance the following Equations: a) ZnCl: + NaOH Zn(OH);+ NaCl c) Al + HCl AICb + H: 6. A 1.3s-g sample of a substance containing carbon, hydrogen, nitrogen and oxygen was burned to produce 0.810 g H:0 and 1.32 g CO. In a separate reaction, all the nitrogen in 0.735 g of the substance was converted to 0.284 g NH. Determine the empirical formula of the substance. If the molar mass of this compound is 65.06g, what is the molecular formula? Oxyacetylene torches used for welding reach a temperature near 2000·The reaction involved is the combustion of acetylene, C2H2 7, 2C2H2(g) + 5O2(g) 4C02(g)+ 2H,01) Starting with175 g of both acetylene and oxygen, what is the theoretical yield, in grams of carbon dioxide? If S2.5 L of carbon dioxide (density - 1.80 gL are produced, what if the percent yield?

Explanation / Answer

Ans. Moles of CO2 produced = Mass / Molar mass

= 1.32 g / (44.0098 g/ mol)

= 0.02999 mol

Moles of H2O produced = 0.810 g / (18.01528 g/ mol) = 0.04496 mol

# Note that-

I. 1 mol CO2­ contains 1 mol C.

Hence, number of moles of C in sample = 0.02999 mol

II. 1 mol H2O contains 2 mol H-atoms.

So,

Moles of H in sample = 2 x moles of H2O produced

                                    = 2 x 0.04496 mol

                                    = 0.08992 mol

Now,

Molar ration of H : C in sample = Moles of H / Moles of C

                                    = 0.08992 mol / 0.02999 mol

                                    = 2.998 : 1

                                    = 3 : 1            (nearest whole number)

That is, there is 3 moles of H for every mol of C.

So,

Empirical formula of compound with respect to C and H = CH3

# Moles NH3 produced from 0.735 g sample = 0.284 g / (17.03056 / mol)

                                                = 0.01668 mol

Since 1 mol NH3 has 1 mol N-atom, the total number of moles of N-atom in 0.735 g sample is equal to moles of NH3 produced.

So, moles of N-atom in 0.735 g sample = 0.01668 mol

Now,

            Moles of N in 1.35 g sample = (1.35 / 0.735) x 0.01668 mol = 0.03063 mol

# Molar ration of N : C in sample = Moles of N / Moles of C

                                    = 0.03063 mol / 0.02999 mol

                                    = 1.02 : 1

                                    = 1 : 1            (nearest whole number)

That is, there is 1 moles of N for every mol of C.

So,

Empirical formula of compound with respect to C and N = CN

# Note that the combustion of 1 mol compound will produce 1 mol CO2 because there is 1 mol C per mol compound.

Thus,

Moles compound in 1.35 g sample = moles of CO­2 produced = 0.02999 mol

Mass of C in 1.35 g compound = Moles of C-atoms in compound x atomic mass of C

= 0.02999 mol x (12.011 g/ mol)

= 0.36025 g

Mass of H in 1.35 g compound = Moles of H-atoms in compound x atomic mass of H

= 0.08992 mol x (1.00794 g/ mol)

= 0.09063 g

Mass of N in 1.35 g compound = Moles of N-atoms in compound x atomic mass of N

= 0.03063 mol x (14.00674 g/ mol)

= 0.42903 g

Mass of O in 1.35 g sample = 1.35 g – (Mass of C + Mass of H + Mass of N)

                                                = 1.35 g – (0.36025 g + 0.09063 g + 0.42903 g)

                                                = 0.47009 g

Moles of O in 1.35 g sample = 0.47009 g / (15.9994 g/ mol)

                                                = 0.02938 mol

# Now,

            Molar ratio of C and O = Moles of O / Moles of C

                                                = 0.02999 mol / 0.02938 mol

                                                = 1.02 : 1

                                                = 1 : 1 (nearest whole number)

That is, there are 1 C-atoms every one O-atm.

So, empirical formula with respect to C and O = CO

## So far we have,

I. Empirical formula of compound with respect to C and H = CH3

II. Empirical formula with respect to C and O = CO

III. Empirical formula with respect to C and N = CN

Comparing statements I, II and III to get a whole number ratio, the empirical formula of compound is (note there must be at least 3 C-atoms to get a whole number ratios) -

                        CH3NO

Hence, empirical formula of the unknown compound = CH3NO

# Molecular weight of empirical formula, CH3NO = 45.04096 g /mol

Ratio of actual and empirical molar mass = 65.06 g mol-1 / 45.04096 g /mol

                                                            = 1.445

Multiply the number of atoms in the empirical formula with the factor 1.445 to get the actual chemical formula – C1.445H4.445N1.445O1.445

Note: Though the number of atoms in chemical formula shall be in while number, the given molar mass and calculated empirical formula does not yield a whole number integers (but a fractional value 1.445). Please check if actual molar mass given is correct.

Because the compound must contain at least 1 atom each of C, H, N, and O – the total mass of these atoms being 43.025 g. To get the whole number ration, the actual molar mass of the compound must be at least multiple of 43.025 or higher than them.

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