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2 years ago

What is the empirical formula of a compound that contains 27.0% s, 13.4% o, and 59.6% cl by mass?

Chemistry

2 answers:

Komok [63]2 years ago

7 0

Answer : The empirical formula of a compound is, $SOCl_2$

Solution :

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of S = 27.0 g

Mass of O = 13.4 g

Mass of Cl = 59.6 g

Molar mass of S = 32 g/mole

Molar mass of O = 16 g/mole

Molar mass of Cl = 35.5 g/mole

Step 1 : convert given masses into moles.

Moles of S = $\frac{\text{ given mass of S}}{\text{ molar mass of S}}= \frac{27.0g}{32g/mole}=0.844moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{13.4g}{16g/mole}=0.837moles$

Moles of Cl = $\frac{\text{ given mass of Cl}}{\text{ molar mass of Cl}}= \frac{59.6g}{35.5g/mole}=1.68moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For S = $\frac{0.844}{0.837}=1.00\aprrox 1$

For O = $\frac{0.837}{0.837}=1$

For Cl = $\frac{1.68}{0.837}=2.00\approx 2$

The ratio of S : O : Cl = 1 : 1 : 2

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = $S_1O_1Cl_2=SOCl_2$

Therefore, the empirical of the compound is, $SOCl_2$

Bezzdna [24]2 years ago

6 0

27g of 's × 1 mole/32.065g equals 0.84
13.4g of o × 1 mole/16g equals 0.83
Divide all these by the smallest number then round to the nearest number.

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In run 1, you mix 7.9 mL of the 43 g/L MO solution (MO molar mass is 327.33 g/mol), 3.13 mL of the 0.040 M SnCl2 in 2.0 M HCl so

blsea [12.9K]

Answer:

Concentration of H3O⁺ [H3O⁺] = 0.864 M

Explanation:

Given that:

The mass concentration of MO = 43 g/L

The volume of MO = 7.9 mL = 7.9 × 10⁻³ L

Recall that

The mass number of MO = Mass concentration of MO × Volume of MO

The mass number of MO = (43 g/L) * (7.9 × 10⁻³ L)

The mass number of MO = 0.3397 g

number of moles of MO = (mass number of MO) / (molar mass of MO)

number of moles of MO = (0.3397 g) / (327.33 g/mol)

moles of MO = 0.00104 mol

The total volume = 7.9 mL + 3.13 mL + 5.49 mL + 3.43 mL

The total volume = 19.95 mL = 19.95 × 10⁻³ L

Concentration of MO [MO} =(number of moles of MO) / (total volume)

[MO] = 0.00104 mol / 19.95 × 10⁻³ L

[MO] = 5.2130 × 10⁻⁸ M

the number of moles of H3O⁺ = molarity of HCl in the solution × the volume of HCl in solution

the number of moles of H3O⁺ = [(2.0 M) * (3.13 mL)] + [(2.0 M) * (5.49 mL)]

the number of moles of H3O⁺ = 17.24 mmol

Concentration of H3O⁺ [H3O⁺] = (the number of moles of H3O⁺) / (total volume)

Concentration of H3O⁺ [H3O⁺] = (17.24 mmol) / (19.95 mL)

Concentration of H3O⁺ [H3O⁺] = 0.864 M

5 0

2 years ago

A rigid, 28-L steam cooker is arranged with a pressure relief valve set to release vapor and maintain the pressure once the pres

MaRussiya [10]

Answer:

$\Delta S_{source}>-1.204\frac{kJ}{K}$

Explanation:

Hello!

In this case, given the initial conditions, we first use the 10-% quality to compute the initial entropy:

$s_1=s_{f,175kPa}+q*s_{fg,175kPa}\\\\s_1=1.4850\frac{kJ}{kg*K} +0.1*5.6865\frac{kJ}{kg*K}=2.0537\frac{kJ}{kg*K}$

Now the entropy at the final state given the new 40-% quality:

$s_2=s_{f,150kPa}+q*s_{fg,150kPa}\\\\s_2=1.4337\frac{kJ}{kg*K} +0.4*5.7894\frac{kJ}{kg*K}=3.7495\frac{kJ}{kg*K}$

Next step is to compute the mass of steam given the specific volume of steam at 175 kPa and the 10% quality:

$m_1=\frac{0.028m^3}{(0.001057+0.1*1.002643)\frac{m^3}{kg} } =0.274kg\\\\m_2=\frac{0.028m^3}{(0.001053+0.4*1.158347)\frac{m^3}{kg} } =0.0603kg$

Then, we can write the entropy balance:

$\Delta S_{source}+\frac{Q}{T_1} -\frac{Q}{T_2} +s_2m_2-s_1m_1-s_{fg}(m_2-m_1)>0$

Whereas sfg stands for the entropy of the leaving steam to hold the pressure at 150 kPa and must be greater than 0; thus we plug in:

Which is such minimum entropy change of the heat-supplying source.

Best regards!

3 0

2 years ago

How much heat is required to raise the temperature of 670g of water from 25.7"C to 66,0°C? The specific heat

inna [77]

Answer:

Explanation:

q= mc theta

where,

Q = heat gained

m = mass of the substance = 670g

c = heat capacity of water= 4.1 J/g°C

theta =Change in temperature=( 66-25.7)

Now put all the given values in the above formula, we get the amount of heat needed.

q= mctheta

q=670*4.1*(66-25.7)

=670*4.1*40.3

=110704.1

8 0

2 years ago

When C2H5Cl(g) is burned in oxygen, chlorine gas is produced in addition to carbon dioxide and water vapor. 5145 kJ of heat are

Arisa [49]

<u>Answer:</u> The chemical equation is written below.

<u>Explanation:</u>

Combustion reaction is defined as the chemical reaction in which a hydrocarbon reacts with oxygen gas to produce carbon dioxide gas and water molecule.

$\text{hydrocarbon}+O_2\rightarrow CO_2+H_2O$

The chemical equation for the combustion of ethyl chloride follows:

$4C_2H_5Cl+13O_2\rightarrow 2Cl_2+8CO_2+10H_2O$

We are given:

When 4 moles of ethyl chloride is burnt, 5145 kJ of heat is released.

For an endothermic reaction, heat is getting absorbed during a chemical reaction and is written on the reactant side.

$A+\text{heat}\rightleftharpoons B$

For an exothermic reaction, heat is getting released during a chemical reaction and is written on the product side

$A\rightleftharpoons B+\text{heat}$

So, the chemical equation follows:

$4C_2H_5Cl+13O_2\rightarrow 2Cl_2+8CO_2+10H_2O+5145kJ$

Hence, the chemical equation is written above.

7 0

2 years ago

Using your data above, draw conclusions about the d-splitting for each ligand (H2O, en, phen). Order the complexes from least to

Delvig [45]

Answer:

H2O<en<phen

Explanation:

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2 years ago