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konstantin123 [22]
2 years ago
8

Mercury chloride is a commercial fungicide. If the molar mass is 470 g/mol and the percent composition is 85.0% Hg and 15.0% Cl,

what is the empirical and molecular formula of the fungicide?
Chemistry
1 answer:
ElenaW [278]2 years ago
6 0
Data: molar mass 470 g/mol

Percent composition:

Hg = 85.0%
Cl = 15.0%

Solution:

1)  Convert % to molar ratios

A. Base: 100 g

=> Hg = 85.0 g / 200.59 g/mol = 0.4235 mol

     Cl = 15.0 g / 35.45 g/mol = 0.4231 mol

B. divide by the higher number and round to whole number

Hg = 0.4325 / 0.4231 = 1.00

Cl = 0.4231 / 0.4231 = 1.00

=> Empirical formula = Hg Cl

2) Find the mass of the empirical formula:

HgCl: 200.59 g/mol + 35.45 g/mol = 236.04

3) Determine how many times is the empirical mass contained in the molecular mass:

470 g/mol / 236.04 = 1.99 ≈ 2

=> Molecular formula = Hg2 Cl2.

Answers:

Empirical formula HgCl
Molecular Formula Hg2Cl2
You might be interested in
How many grams of NH3 can be prepared from the synthesis of 77.3 grams of nitrogen and 14.2 grams of hydrogen gas?
lbvjy [14]

Answer:

80.41 g

Explanation:

Data Given:

Mass of Nitrogen (N₂) = 77.3 g

Mass of Hydrogen (H₂) = 14.2 g

many grams of NH₃ = ?

Solution:

First we look at the balanced synthesis reaction

              N₂   +    3 H₂  ------—> 2 NH₃

             1 mol      3 mol

As 1 mole of Nitrogen react with 3 mole of hydrogen

Convert moles to mass

molar mass of N₂ = 2(14) = 28 g/mol

molar mass of H₂ = 2(1) + 2 g/mol

Now

                     N₂             +           3 H₂        ------—>      2 NH₃

             1 mol (28 g/mol)     3 mol(2g/mol)

                    28 g                        6 g

28 grams of N₂ react with 6 g of H₂  

So

if 28 grams of N₂ produces 6 g of H₂  so how many grams of N₂ will react with 14.2 g of H₂.

Apply Unity Formula

                 28 g of N₂ ≅ 6 g of H₂

                 X g of N₂ ≅ 14.2 g of H₂

Do cross multiply

                X g of N₂ = 28 g x 14.2 g / 6 g

                X g of N₂ = 66.3 g

As we have given with 77. 3 g of N₂ but from this calculation we come to know that 66.3 g will react with 14.2 g of hydrogen and the remaining 10 g N₂ will be in excess

So, Hydrogen is limiting reactant in this reaction and the amount of NH₃ depends on the amount of hydrogen.

Now

To find mass of NH₃ we will do following calculation

Look at the reaction

As we Know

                     N₂             +           3 H₂        ------—>      2 NH₃

                                                   6 g                            2 mol

So, 6 g of hydrogen gives 2 moles of NH₃, then how many moles of NH₃ will be produce by 14.2 g

Apply Unity Formula

                 6 g of H₂ ≅ 2 mol of NH₃

                14.2 g of H₂ ≅ X mol of NH₃

Do cross multiply

               X mol of NH₃= 14.2 g x 2 mol / 6 g

                X mol of NH₃ = 4.73 mol

So, 14.2 g of hydrogen gives 4.73 moles of NH₃

Now

Convert moles of NH₃ to mass

Formula will be used

        mass in grams = no. of moles x molar mass . . . . . . (2)

Molar mass of  NH₃

Molar mass of  NH₃ = 14 + 3(1)

Molar mass of  NH₃ = 14 + 3 = 17 g/mol

Put values in equation 2

        mass in grams = 4.73 mole x 17 g/mol

        mass in grams =  80.41 g

mass of NH₃=  80.41 g

3 0
2 years ago
3. According to the label on a bottle of concentrated hydrochloric acid, the contents are 36.0% HCl by mass and have a density o
velikii [3]

Answer:

a) 11.64 M

b) 43 mL

c) 1.7 kg

Explanation:

a) Let's use a basis of the calculus of 1000 mL (1 L) of the concentrated solution. If the solution has 1.18 g/mL, it has:

1.18*1000 = 1180 g.

The mass of HCl will be then:

mHCl = 1180*0.36 = 424.8 g

The molar mass of HCl is 36.5 g/mol, so the number of moles is the mass divided by the molar mass:

nHCl = 424.8/36.5 = 11.64 mol

The molarity is the number of moles divided by the volume in L:

Molarity = 11.64 M

b) To prepare a solution by dilution of a concentrated one, we can use the equation:

C1V1 = C2V2

Where C is the concentration, V is the volume, 1 is the concentrated solution, and 2 the final solution. So:

11.64*V1 = 2.00*0.250

V1 = 0.0429 L ≅ 43 mL

c) The neutralization will happen by the equation:

HCl + NaHCO₃ → NaCl + CO₂ + H₂O

So, 1 mol of NaHCO₃ is needed to react with 1 mol of HCl. At 1.75 L, the number of moles of the acid is:

nHCl = 1.75*11.64 = 20.37 mol

The molar mass of NaHCO₃ is 84 g/mol so the mass needed is the molar mass multiplied by the number of moles:

m = 84*20.37 = 1,711.08 g

m = 1.7 kg

6 0
2 years ago
vanadium has an atomic mass of 50.9415 amu. it has two common isotopes.one isotopes has a mass of 50.9440 amu and a relative abu
Kaylis [27]

Explanation:

Average atomic mass of the  vanadium = 50.9415 amu

Isotope (I) of vanadium' s abundance = 99.75 %= 0.9975

Atomic mass of Isotope (I) of vanadium ,m= 50.9440 amu

Isotope (II) of vanadium' s abundance =(100%- 99.75 %) = 0.25 % = 0.0025

Atomic mass of Isotope (II) of vanadium ,m' = ?

Average atomic mass of vanadium =

m × abundance of isotope(I) + m' × abundance of isotope (II)

50.9415 amu =50.9440 amu× 0.9975 + m' × 0.0025

m'= 49.944 amu

The atomic mass of isotope (II) of vanadium is 49.944 amu.

5 0
2 years ago
Heavy nuclides with too few neutrons to be in the band of stability are most likely to decay by what mode?
Margarita [4]

Answer:

Positron emission

Explanation:

Positron emission involves the conversion of a proton to a neutron. This process increases the mass number of the daughter nucleus by 1 while its atomic number remains the same. The new neutron increases the number of neutrons present in the daughter nucleus hence the process increases the N/P ratio.

A positron is usually ejected in the process together with an anti-neutrino to balance the spins.

6 0
2 years ago
When 10 g of diethyl ether is converted to vapor at its boiling point, about how much heat is absorbed? (c4h10o, δhvap = 15.7 kj
vagabundo [1.1K]

Answer:

ΔH= 3KJ

Explanation:

The total heat absorbed is the total energy in the process, and that is in form of entalpy.

ΔH = q + ΔHvap, where q is the heat necessary for elevate the temperature of dietil ether. Suppose the initial temperature is room temperature (25ºC=298 K), then

q= 10g x2.261 J/gK x(310 K - 298K)= 271.32 J= 0.3 kJ

Then

ΔHvap = 10g C4H10O x (1 mol C4H10O/74.12 g C4H10O) x( 15.7 KJ/ 1 mol C4H10O) = 2.12 KJ

ΔH= 2.5KJ ≈ 3KJ

5 0
2 years ago
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