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Answer to Question #163275 in Chemical Engineering for Neema

Question #163275

Two of the most common rocks at the surface of the earth are basalt and granite. As a budding engineering student, you would like to find the ratio of basalt to granite in a rock sample which has an average density of 2.96 g/cm3. The granite in the sample has a volume of 1.00 ft3 and weighs 171 lbs. The piece of basalt has a volume of 64 in3 and weighs 116 ounces. Find the percentage of basalt and granite which give the sample a density of 2.96 g/cm3

1
2021-02-16T07:29:19-0500

Q163275

Two of the most common rocks at the surface of the earth are basalt and granite. As a budding engineering student, you would like to find the ratio of basalt to granite in a rock sample that has an average density of 2.96 g/cm3. The granite in the sample has a volume of 1.00 ft3 and weighs 171 lbs. The piece of basalt has a volume of 64 in3 and weighs 116 ounces. Find the percentage of basalt and granite which give the sample a density of 2.96 g/cm3.

Solution:

All the quantities given in the question are in different units. We will use the same units for mass and volume so that our calculation becomes easy.

For simplicity, it is better for us to use mass in 'kg' and volume in 'L'.

mass and volume of granite in the sample is already given to us.

Step 1: Convert the mass of granite to 'kg' units and volume to 'L' units.

Granite :

mass = 171lbs.

Volume = 1.00 ft3 .

Convert 171 lbs by using the conversion factor, 1 kg = 2.2046 lbs.

"mass\\space in \\space 'kg' = 171\\space lbs * \\frac{1kg}{2.2046lbs} = 77.564 \\space kg"

Convert 1.00 ft3 to 'L' by using the conversion factors

1ft = 12inch, 1inch = 2.54 cm , 1L = 1000 cm3.

"volume\\space in \\space 'L' = 1.00ft^{3}* (\\frac{12inch}{1ft})^{3}*(\\frac{2.54cm}{1inch})^{3}*\\frac{1L}{1000cm^{3}}"

"= 1.00ft^{3}* \\frac{1728inch^{3}}{1ft^{3}}*\\frac{16.387cm^{3}}{1inch^{3}}*\\frac{1L}{1000cm^{3}}"

"=\\frac{1728*16.387}{1000}L = 28.3167 L"

Step 2 :

We are not provided the mass and volume of basalt in the rock sample. Instead, we are given the mass and volume of a sample of pure basalt. Using this we can find the density of the basalt.

Convert 116 ounces to 'kg' by using the conversion factor

1kg = 2.2046lbs and 1lbs = 16 ounces

"mass \\space of \\space basalt\\space in\\space 'kg' = 116ounces * \\frac{1lbs }{16ounce }*\\frac{1kg}{2.2046lbs } = 3.28854\\space kg"

Convert 64 inch3 to 'L' by using the conversion factor

1inch = 2.54cm . and 1L = 1000 cm3.

"volume\\space in\\space 'L' = 64inch^{3} * (\\frac{2.54cm}{1inch})^{3}*\\frac{1L}{1000cm^{3}}"

"=64inch^{3}* \\frac{16.387cm^{3}}{1inch^{3}}*\\frac{1L}{1000cm^{3}} = 1.04877\\space L"

We need the mass and volume of basalt only for finding the density of basalt.

"density = \\frac{mass}{volume} = \\frac{3.28854\\space kg}{1.04877\\space L} = 3.1356\\space kg\/L"

Step 3: Convert density of rock sample to kg/L

convert 2.96 g/cm3 to 'kg/L ' by using the conversion factor

1kg = 1000g and 1L = 1000cm3 .

"the \\space density\\space of\\space rock\\space sample\\space in\\space 'kg\/L' = \\frac{2.96\\space g}{1\\space cm^{3}}*\\frac{1000\\space cm^{3}}{1\\space L}*\\frac{1\\space kg}{1000\\space g}= 2.96 kg\/L"

Step 4 :

Let us assume that the mass of the rock sample is 'x'.

We already know the mass of granite in the sample which is = 77.564 kg.

So the mass of basalt in the rock sample = ( x - 77.564 )

We will use the volume additive equation here.

"volume\\space of \\space granite + volume\\space of\\space basalt = volume\\space of\\space the\\space rock\\space sample."

"volume \\space of \\space granite\\space in\\space the\\space sample\\space is\\space 28.3167 L."

"volume \\space of\\space basalt =\\frac{ mass}{ density } =\\frac{ (x - 77.564)}{3.1356 kg\/L}" ;

"volume \\space of\\space rock\\space sample =\\frac{ mass}{ density } =\\frac{x}{2.96\\space kg\/L}"

plug all this in the equation of volume addition we have

"28.3167L * \\frac{ (x - 77.564)}{3.1356 kg\/L} =\\frac{x}{2.96\\space kg\/L}"

88.7907 kg + ( x - 77.564) = 1.0593 x

88.7907 - 77.564 = 1.0593 x - x ;

11.2264 kg = 0.0593 x

x = 11.2264kg / 0.0593 = 189.32 kilograms.

Hence the mass of the rock sample = 189.32 kilograms.

mass of granite = 77.564 kilograms

and mass of basalt in the rock sample = 189.32 - 77.564 = 111.756 kilograms.

Step 5: To find the % by weight of granite and basalt in the rock sample

% by mass of granite = 77.564kg / 189.32kg * 100 = 40.97 %

% by mass of basalt = 111.756 kg/189.32kg * 100 = 59.03%

"% by weight of granite = \\frac{mass of }{b}= \\frac{77.564 kg }{189.32 kg } * 100 =" "% \\space by\\space mass\\space of\\space granite = \\frac{mass \\space of\\space granite }{mass \\space of\\space rock\\space sample }*100=\\frac{77.564\\space kg}{189.32\\space kg} *100 = 40.97%"

So when the % of granite is 40.97% and % of basalt is 59.03% in the rock sample the density of the rock sample we get is 2.96g/cm^3

Please let me know if you have not understood any step.

Thank you. :)

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