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loris [4]
2 years ago
15

A 0.8715 g sample of sorbic acid, a compound first obtained from the berries of a certain ash tree, is burned completely in oxyg

en to give 2.053 g of carbon dioxide and 0.5601 g of water. the empirical formula of sorbic acid is
Chemistry
2 answers:
aivan3 [116]2 years ago
8 0
<h3><u>Answer</u>;</h3>

C3H4O

<h3><u>Explanation;</u></h3>

Empirical formula is the simplest formula of a compound;

Molar mass CO2 = 44.01  

Mass of CO2 produced = 2.053 g

Mass of carbon in original sample = 12.01/44.01 × 2.053

                                                           = 0.5603g  

Molar mass H2O = 18  

Mass of H in original sample = 2/18 ×0.5601

                                               = 0.0622  g

Thus; original sample contained 0.5603g C and 0.0622g H. The balance of the sample was O  

Mass of O = 0.8715 - (0.5603 + 0.0622) = 0.249g  

The mole ratio of C:H:O  will be;

Moles C = 0.5603/12 = 0.0467  

Moles H = 0.0622  

Moles O = 0.249/16 = 0.01556  

C:H:O = 0.0467:0.0622:0.01556  

Divide through by 0.01556:  

C:H:O = 3:4:1  

Empirical formula is thus C3H4O

agasfer [191]2 years ago
4 0

Answer:

C₃H₄O is the empirical formula of ascorbic acid

Explanation:

First, we need to write the reaction. I will call the Ascorbic Acid as AA:

AA + O₂ ------> CO₂ + H₂O

We know we have initially 0.8715 g of AA, and this produces 2.053 g of CO₂ and 0.5601 g of H₂O.

From the products, we can calculate how many moles of each element (C, H and O) has the ascorbic acid.

So, let's take in mind the molecular weight of CO₂ and H₂O:

MWCO₂ = 44 g/mol

MWH₂O = 18 g/mol

From the molecular weight we can calculate the moles of each product:

moles CO₂ = 2.053 / 44 = 0.0467 moles

moles H₂O = 0.5601 / 18 = 0.0311 moles

Now, with this, we can do a mole ratio of the elements. In the case of CO₂, we can calculate how many grams of Carbon are in the ascorbic acid, and with the water, the grams of hydrogen. Then, with difference we can get the mass of oxygen.

To do this ratio, we do a rule of 3.

To Carbon:

12 g C --------> 44 g CO₂

X g C ---------> 2.053 g CO₂

X = 2.053 * 12/44 = 0.5599 g of C

For Hydrogen:

2 g H ------> 18 g H₂O

X g H ------> 0.5601 g H₂O

X = 2 * 0.5601/18 = 0.0622 g of H

We have the mass of C and H, so the grams of oxygen are:

m O = 0.8715 - 0.0622 - 0.5599 = 0.2494 g of O

Now that we have the mass of each element, we can calculate the empirical formula following 3 steps.

Step 1: Calculate the moles of each element

C = 0.5599 / 12 = 0.0467

H = 0.0622 / 1 = 0.0622

O = 0.2494 / 16 = 0.0156

Step 2: Calculate the mole ratio

Doing this, is just dividing all the moles by the smallest number of moles obtained. In this case, the oxygen so:

C = 0.0467/0.0156 = 2.99

H = 0.0622/0.0156 = 3.99

O = 0.0156/0.0156 = 1

These results represent the number of atoms of each element in the empirical formula, so the last step is write the empirical formula with the previous results, rounding the numbers to get an even number so:

C₃H₄O

This would be the empirical formula of ascorbic acid.

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A student is given a sample of a blue copper sulfate hydrate. He weighs the sample in a dry covered porcelain crucible and got a
Nata [24]

Answer:

There are present 5,5668 moles of water per mole of CuSO₄.

Explanation:

The mass of CuSO₄ anhydrous is:

23,403g - 22,652g = 0,751g.

mass of crucible+lid+CuSO₄ - mass of crucible+lid

As molar mass of CuSO₄ is 159,609g/mol. The moles are:

0,751g ×\frac{1mol}{159,609g} = 4,7052x10⁻³ moles CuSO₄

Now, the mass of water present in the initial sample is:

23,875g - 0,751g - 22,652g = 0,472g.

mass of crucible+lid+CuSO₄hydrate - CuSO₄ - mass of crucible+lid

As molar mass of H₂O is 18,02g/mol. The moles are:

0,472g ×\frac{1mol}{18,02g} = 2,6193x10⁻² moles H₂O

The ratio of moles H₂O:CuSO₄ is:

2,6193x10⁻² moles H₂O / 4,7052x10⁻³ moles CuSO₄ = 5,5668

That means that you have <em>5,5668 moles of water per mole of CuSO₄.</em>

I hope it helps!

5 0
2 years ago
A gas cylinder filled with nitrogen at standard temperature and pressure has a mass of 37.289 g. The same container filled with
andrew-mc [135]

Answer:

Molar mass = 3.9236 g/mol ≅ 4 g/mol

This corresponds to Helium gas.

Explanation:

Let the moles of nitrogen gas = x moles

Moles of carbon dioxide = x moles ( As both are filled at same temperature and pressure conditions )

Given:

Mass_{Container}+Mass_{Nitrogen\ gas}=37.289\ g

Molar mass of nitrogen gas, N_2 = 28.014 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

x\ moles= \frac{Mass}{28.014\ g/mol}

Mass of nitrogen gas = 28.014x g

So,

Let, Mass_{Container}=y

y+28.014x=37.289

Similarly,

Mass_{Container}+Mass_{Carbon\ dioxide\ gas}=37.440\ g

Molar mass of nitrogen gas, CO_2 = 44.01 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

x\ moles= \frac{Mass{44.01\ g/mol}

Mass of nitrogen gas = 44.01x g

So,

y+44.01x=37.440

Solving the two equations, we get :

Mass_{Container}=y=37.025\ g

x = 0.00943 moles

Thus, Given:

Mass_{Container}+Mass_{Unknown\ gas}=37.062\ g

37.025\ g+Mass_{Unknown\ gas}=37.062\ g

Mass of the gas = 0.037 moles

Moles = 0.00943 moles

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

0.00943\ moles= \frac{0.037\ g}Molar mass}

Molar mass = 3.9236 g/mol ≅ 4 g/mol

This corresponds to Helium gas.

7 0
2 years ago
A Mole of elephants would weigh how much more than the moon? I got 6.301x10^23 and set the problem up like this: 6.022x10^23 x 2
Margaret [11]

Answer:

The Moon weighs about 7.349*10^22 kg

How much does “an elephant” weigh? Grown-up elephants vary between 1.7 tons (small wood elephant cow) and 6 tons (big African elephant bull). Let us pick 2.720 tons or 2720 kg as an average weigh one, this is valid for Asian elephants.

A mole of elephants weighs 6.022 * 10^23 x 2720 kg = 1.638 * 10^27 kg.

That is 1,638 * 10^27 / 7,349*10^22 = 22289 times more.

5 0
2 years ago
What is the kinetic energy acquired by the electron in hydrogen atom, if it absorbs a light radiation of energy 1.08x101 J. (A)
Delvig [45]

Explanation:

The given data is as follows.

            Energy of radiation absorbed by the electron in hydrogen atom = 1.08 \times 10^{-17} J

As energy is absorbed as a photon. Hence, frequency will be calculated will be as follows.

                                    E = h \nu

               1.08 \times 10^{-17} J = 6.626 \times 10^{-34} Js \times \nu

               \nu = 0.163 \times 10^{17} s^{-1}

or,                \nu = 1.63 \times 10^{16} s^{-1}    

It is known that,        \nu = \frac{c}{\lambda}

                1.63 \times 10^{16} s^{-1} = \frac{3 \times 10^{8} m/s}{\lambda}                  

                   \lambda = 1.84 \times 10^{-8} m

And, according to De-Broglie equation \lambda = \frac{h}{p}

as,        p = m \times \nu

So,          \lambda = \frac{h}{m \times \nu}

            m \times \nu = \frac{6.626 \times 10^{-34} Js}{1.84 \times 10^{-8} m}          

                             = 3.6 \times 10^{-26} J/m

Now, on squaring both the sides we get the following.

           (m \times \nu)^{2} = (3.6 \times 10^{-26} J/m)^{2}    

                              = 12.96 \times 10^{-52}  

               m \times \nu^{2} = \frac{12.96 \times 10^{-52}}{m}

where,   m = mass of electron

So,           m \times \nu^{2} = \frac{12.96 \times 10^{-52}}{m}

                             = \frac{12.96 \times 10^{-52}}{9.1 \times 10^{-31}}

                                   = 1.42 \times 10^{-21} J

Since,  K.E = \frac{1}{2}m \nu^{2}

                 = \frac{1.42 \times 10^{-21} J}{2}

                 = 0.71 \times 10^{-21} J

Thus, we can conclude that kinetic energy acquired by the electron in hydrogen atom is 7.1 \times 10^{-22} J.

4 0
2 years ago
The piece of iron that miguel measured had a mass of 51.1 g and a volume of 6.63 cm 3 . what did miguel calculate to be the dens
likoan [24]
Given:
Mass, m = 51.1 g
Volume, V = 6.63 cm³

By definition, 
Density = Mass/Volume
              = (51.1 g)/(6.63 cm³)
              = 7.7074 g/cm³

In SI units,
Density = (7.7074 g/cm³)*(10⁻³ kg/g)*(10² cm/m)³
              = 7707.4 kg/m³

Answer: 7.707 g/cm³ or 7707.4 kg/m³

4 0
2 years ago
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