# Tutor profile: Samantha P.

## Questions

### Subject: Chemistry

How much carbon dioxide is produced by burning 5.00g of ethane?

Step 1: Write a chemical equation to describe the reaction in the question. (Hint: Burning is a combustion reaction, meaning that a fuel is reacting with oxygen to produce carbon dioxide and water): $$C_2H_6$$ + $$O_2$$ --> $$CO_2$$ + $$H_2O$$ Step 2: Balance the chemical equation by adding coefficients: 2$$C_2H_6$$ + 7$$O_2$$ --> 4$$CO_2$$ + 6$$H_2O$$ From the chemical equation, we can see that 4 mol of $$CO_2$$ is produced for every 2 mol of $$C_2H_6$$ burned (assuming that oxygen is present in excess). Step 3: Find (or better - calculate) the molecular weight of ethane and of carbon dioxide: MW (ethane) = (2 x 12.011 g/mol) + (6 x 1.008 g/mol) = 30.07 g/mol MW (carbon dioxide) = 12.011 g/mol + (2 x 15.999 g/mol) = 44.01 g/mol Step 4: Calculate the number of moles of ethane: n ($$C_2H_6$$) = $$\frac{m}{MW}$$ n ($$C_2H_6$$) = $$\frac{5.00 g}{30.07 g/mol}$$ n ($$C_2H_6$$) = 0.166 mol Step 5: Calculate the number of moles of carbon dioxide: Remember, for every 2mol of $$C_2H_6$$, we will produce 4mol of $$CO_2$$, so: n ($$CO_2$$) = $$\frac{4}{2}$$ x n ($$C_2H_6$$) n ($$CO_2$$) = $$\frac{4}{2}$$ x 0.166 mol n ($$CO_2$$) = 0.332 mol Step 6: Calculate the mass of carbon dioxide from the moles of carbon dioxide: m = n x MW m = 0.332 mol x 44.01 g/mol m = 14.61 g Therefore, by burning 5.00 g of ethane, 14.61 g of carbon dioxide is produced.

### Subject: Biochemistry

How many grams of monosodium phosphate ($$ NaH_{2}PO_{4} $$, MW = 119.98g/mol, pKa = 6.82) and disodium phosphate ($$ Na_{2}HPO_{4} $$, MW = 177.99g/mol) are required to make 0.5L of a 0.5M buffer with a pH of 7.1 ?

Notice that monosodium phosphate and disodium phosphate are a conjugate acid and base pair. This means that the dissociation of a hydrogen ion ($$H^{+}$$) from the conjugate acid (monosodium phosphate) results in the formation of the conjugate base (disodium phosphate). This can be demonstrated by writing out the net dissociation equation (notice that sodium is excluded from the net equation since it is a spectator ion): $$H_{2}PO_{4}^{-}$$ <---> $$H^{+}$$ + $$H_{2}PO_{4}^{2-} $$ Since we have a conjugate acid and base pair, we can use the Henderson-Hasselbach equation: pH = pKa + $$log_{10}$$$$ \frac{[Base]}{[Acid]}$$ If we rearrange this equation, we get: $$\frac{[Base]}{[Acid]}$$ = $$10^{pH-pKa}$$ Plug in the pH and pKa that we were given in the question: $$\frac{[Base]}{[Acid]}$$ = $$10^{7.1-6.82}$$ = 1.905 Now we can isolate either the concentration of the acid or the concentration of the base. For this example, let's isolate the base: [Base} = 1.905 {Acid} We also know from the question that the molarity of the solution will be 0.5M, so: [Acid] + [Base] = 0.5 mol/L Since we have already defined {Base], we can replace this with [Base} = 1.905 {Acid}, which gives us: [Acid] + 1.905 {Acid} = 0.5 mol/L Which simplifies to: 2.905 {Acid} = 0.5 mol/L Solving for acid gives us: [Acid] = $$\frac{0.5mol/L}{2.905}$$ = 0.172 mol/L Recall that [Base} = 1.905 {Acid}. Therefore, [Base} = 1.905 x 0.172 mol/L = 0.328 mol/L Now that we have the concentrations of the conjugate acid and base, we can calculate the number of moles, then the mass needed to produce the desired buffer. Let's start with the acid: n = C x V n = 0.172mol/L x 0.5L n = 0.086 mol m = n x MW m = 0.086 mol x 119.98 g/nol m = 10.32 g And then the same for the base: n = C x V n = 0.328mol/L x 0.5L n = 0.164 mol m = n x MW m = 0.164 mol x 177.99 g/nol m = 29.19 g Therefore, to make 0.5L of buffer at pH 7.1, we will need 10.32 g of monosodium phosphate and 29.19 g of disodium phosphate.

### Subject: Basic Chemistry

How would you name the following chemical compound: $$CF_{4}$$ ?

Step 1: Identify the elements (consult periodic table if necessary) C = carbon and F = fluorine Step 2: Determine if the compound is ionic or molecular Since carbon (C) and fluorine (F) are both non-metals (found on the top right of the periodic table) they must form a molecular bond. Therefore, we must use molecular naming conventions. Step 3: Apply molecular naming convention The second element's ending is changed to "-ide". Hence fluorine becomes fluoride. (Note that naming conventions usually name the element with the lowest electronegativity - in this case carbon - first) Add prefixes to indicate the number of each element (denoted by the subscript) one = "mono-" and four = "tetra-". However, "mono" is usually implied in the absence of any other prefix. Hence, the answer is carbon tetrafluoride.

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