How do I solve this chemistry question?

Question

Hi, I have tried so many methods but I cannot figure out the steps or the answer of this to save my life Dx.

What mass (in mg) of sodium benzoate (C_7 H_5 O_2 Na) is present in 4.56 x 10^2 mL of a 0.345 M solution?

I mainly had trouble converting from moles to milligrams. Thank you soo much!!

Answer 1

definitely follow the advice of pisgahchemist.

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you need to learn and apply "factor label method".. aka.. "unit factor method"... aka.. "dimensional analysis (in chemistry class)
I cover the basics in my answer here
https://answers.yahoo.com/question/index...
spend 10 min and read through that. It will change your life.

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this problem

we're converting FROM: ... ... ... ... ... .4.56x10^2 mL of solution
we're converting TO: .. ... ... ... .. ... .. . mg C7H5O2Na
we're given this UNIT FACTOR:.. .. ... 0.345 mol C7H5O2Na / L solution

and we'll need this additional unit factor.... 144.11g C7H5O2Na / 1 mol C7H5O2Na

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then we start with the FROM on the left and convert to the TO on the right

. 4.56x10^2mL solution... .. 1 L... ..0.345 mol C7H5O2Na.. 144.11g C7H5O2Na.. 1000mg
---- ---- ----- ---- ---- ----- x ---- ---- x ---- --- --- --- ---- --- --- x ---- ---- ---- ---- ---- --- x ---- ----- = 2.27x10^4 mg C7H5O2Na
.. ... .... ... .1.. ... ... .. ... .. .1000mL.. .. ..1 L solution .. .... . ... 1 mol C7H5O2Na.. . . . 1g

which you can calculate by entering
.. 4.56e2 * 0.345 * 144.11 =
in your calculator.

Answer 2

Chemistry question....

Avoid the hand-waving and simply use the unit-factor method .... as taught by just about every chemistry teacher in the known universe.

456 mL x (1L / 1000 mL) x (0.345 mol C7H5O2Na / 1L) x (144.0g C7H5O2Na / 1 mol C7H5O2Na) x (1000 mg / 1g) = 22,700 mg C7H5CO2Na ............... express the answer to three significant digits

Unit factor method: "One line, one chain calculation, zero errors."

Answer 3

one mole of sodium binzoate weighs 7(12.01 g)+5(1.008g) +2(16g) +1(22.99g) = 144.6 grams where the numbers in parenthesis are the atomic weights of the elements.
One liter of a 0.345M solution contains 0.345 (144.6g)=49.7 grams of sodium benzoate.
Convert 4.56x10^2 mL to liters = 456mL /1000mL/L = 0.456 L
SO, the grams of sodium Benzoate in 0.456L = >456/1000 fraction of a liter x 49.7g =22.663 grams.=22663 mg

Answer 4

To determine the mass of sodium benzoate, multiply the mass of one mole by the number of moles. The first step is to calculate the mass of one mole of sodium benzoate.

Mass = 12 * 7 + 5 + 16 * 2 + 23 = 144 grams

0.345 M
This means one liter of the solution contains 0.345 mole of sodium benzoate. To determine the number of moles of sodium benzoate, multiply the volume liters by this number.

n = 0.456 * 0.345 = 0.15732

Mass of sodium benzoate = 144 * 0.15732 = 22.65408 g
Rounded to three significant digits, the mass is 22.7 grams.

Answer 5

First determine the number of moles:
4.56e2 ml * (0.345 moles/1000 ml) = 0.15732 moles

Then determine the molar mass of NaC7H5O2. You can use a periodic table and add up all the masses, but it's a lot easier to just ask online. https://www.webqc.org/molecular-weight-o...

Molar mass of NaC7H5O2 = 144.1032 g/mole

Now just multiply
0.15732 moles * 144.1032 g/mole = 22.7 grams = 2.27e4 mg