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3 years ago

Empirical formula of 25.13g sulfur and 49.87g copper

Chemistry

1 answer:

antiseptic1488 [7]3 years ago

7 0

Copper + sulfur create copper sulfide.

It's a lengthy process to find the empirical formula:

Sulfur:

$25.13g * \frac{1 mol}{32.065 g} = 0.78 mol$

Copper:

$49.87 * \frac{1 mol}{63.546 g} = 0.78mol$

S:Cu => $\frac{0.78}{0.78} :\frac{0.78}{0.78}$ => $(1:1) * 1$ => $1:1$

The empirical formula would just be $CuS$

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A student weighs an empty flask and stopper and finds the mass to be 55.844 g. She then adds about 5 mL of an unknown liquid and

Oduvanchick [21]

Answer :

(a) The pressure of the vapor in the flask in atm is, 0.989 atm

(b) The temperature of the vapor in the flask in Kelvin is, 372.7 K

The volume of the flask in liters is, 0.2481 L

(c) The mass of vapor present in the flask was, 0.257 g

(d) The number of moles of vapor present are 0.00802 mole.

(e) The mass of one mole of vapor is 32.0 g/mole

Explanation : Given,

Mass of empty flask and stopper = 55.844 g

Volume of liquid = 5 mL

Temperature = $99.7^oC$

Mass of flask and condensed vapor = 56.101 g

Volume of flask = 248.1 mL

Barometric pressure in the laboratory = 752 mmHg

(a) First we have to determine the pressure of the vapor in the flask in atm.

Pressure of the vapor in the flask = Barometric pressure in the laboratory = 752 mmHg

Conversion used :

$1atm=760mmHg$

or,

$1mmHg=\frac{1}{760}atm$

As, $1mmHg=\frac{1}{760}atm$

So, $752mmHg=\frac{752mmHg}{1mmHg}\times \frac{1}{760}atm=0.989atm$

Thus, the pressure of the vapor in the flask in atm is, 0.989 atm

(b) Now we have to determine the temperature of the vapor in the flask in Kelvin.

Conversion used :

$K=273+^oC$

As, $K=273+^oC$

So, $K=273+99.7=372.7$

Thus, the temperature of the vapor in the flask in Kelvin is, 372.7 K

Now we have to determine the volume of the flask in liters.

Conversion used :

1 L = 1000 mL

or,

1 mL = 0.001 L

As, 1 mL = 0.001 L

So, 248.1 mL = 248.1 × 0.001 L = 0.2481 L

Thus, the volume of the flask in liters is, 0.2481 L

(c) Now we have to determine the mass of vapor that was present in the flask.

Mass of flask and condensed vapor = 56.101 g

Mass of empty flask and stopper = 55.844 g

Mass of vapor in flask = Mass of flask and condensed vapor - Mass of empty flask and stopper

Mass of vapor in flask = 56.101 g - 55.844 g

Mass of vapor in flask = 0.257 g

Thus, the mass of vapor present in the flask was, 0.257 g

(d) Now we have to determine the number of moles of vapor present.

Using ideal gas equation:

PV = nRT

where,

P = Pressure of vapor = 0.989 atm

V = Volume of vapor = 0.2481 L

n = number of moles of vapor = ?

R = Gas constant = 0.0821 L.atm/mol.K

T = Temperature of vapor = 372.7 K

Putting values in above equation, we get:

$(0.989atm)\times 0.2481L=n\times (0.0821L.atm/mol.K)\times 372.7K\\\\n=0.00802mole$

Thus, the number of moles of vapor present are 0.00802 mole.

(e) Now we have to determine the mass of one mole of vapor.

$\text{Mass of one mole of vapor}=\frac{\text{Mass of vapor}}{\text{Moles of vapor}}$

$\text{Mass of one mole of vapor}=\frac{0.257g}{0.00802mole}=32.0g/mole$

Thus, the mass of one mole of vapor is 32.0 g/mole

8 0

3 years ago

A metal X from two oxide A and B .3.oogm of A and B contain 0.72 and 1.16g of oxygen respectively.calculate the maases of metal

devlian [24]

Answer:

Explanation:

Firstly, we have to determine the mass of metal X. We can do that by interpreting the first and second statement mathematically.

Metal X can form 2 oxides (A and B).

A + B = 3g

The mass of oxygen in A is 0.72g and the mass of oxygen in B is 1.16g.

The mass of metal X in the two oxides will be the same because it's the same metal.

Thus, we represent the mass of the metal in the two oxides as 2X.

2X + 0.72 + 1.16 = 3

2X + 1.88 = 3

2X = 3 - 1.88

2X = 1.12

X = 0.56

Thus, 0.56 g of the metal combines with 0.72g of oxygen in A and 1.16 g of oxygen in B.

Thus, mass of metal (X) in 1g of oxygen in A is

0.56g ⇒ 0.72g

X ⇒ 1

X = 1 × 0.56/0.72

X = 0.78 g

Hence, 0.78g of the metal will combine with 1g of oxygen for A

Also, mass of metal (X) in 1g of oxygen in B is

0.56g ⇒ 1.16g

X ⇒ 1g

X = 1×0.56/1.16

X = 0.48 g

Thus, 0.48g of the metal will combine with 1g of oxygen for B

6 0

3 years ago

What is the number of atoms in a mole of any element?

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6.022×10^23 should be correct. Are there any options to choose from?

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3 years ago

What is the percent error of a length measurement of .454cm if the correct value is 0.546cm?

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