An analytical sample contains an initial solute mass of $2.45\text{ g}$. During an assay, a chemist withdraws $320\text{ mg}$ of solute, adds two successive aliquots each containing $y\text{ mg}$, and finally removes $0.18\text{ g}$. If the remaining mass of solute in the mixture is exactly $3.65\text{ g}$, solve the resulting multi-step balance equation for $y$ in milligrams.

An analytical sample contains an initial solute mass of $2.45\text{ g}$ . During an assay, a chemist withdraws $320\text{ mg}$ of solute, adds two successive aliquots each containing $y\text{ mg}$ , and finally removes $0.18\text{ g}$ . If the remaining mass of solute in the mixture is exactly $3.65\text{ g}$ , solve the resulting multi-step balance equation for $y$ in milligrams.

Final answer

$y = 850\text{ mg}$

Steps

Convert all quantities to a single unit

To find the value of $y$ , write a mass-balance equation representing the solute before and after the additions and removals.

Since $y$ is expressed in milligrams ( $\text{mg}$ ), convert all masses given in grams ( $\text{g}$ ) to milligrams using the conversion factor $1\text{ g} = 1000\text{ mg}$ :

Initial mass: $2.45\text{ g} = 2.45 \times 1000 = 2450\text{ mg}$
Final mass: $3.65\text{ g} = 3.65 \times 1000 = 3650\text{ mg}$
Mass removed at the end: $0.18\text{ g} = 0.18 \times 1000 = 180\text{ mg}$

Common Pitfall: Mixing units without conversion (e.g., subtracting $320\text{ mg}$ directly from $2.45\text{ g}$ ) leads to an incorrect scale and result.

Set up the mass-balance equation

Track each step of the procedure: $\text{Initial mass} - \text{first withdrawal} + \text{two aliquots} - \text{second withdrawal} = \text{Final mass}$

Substitute the values: $2450 - 320 + 2y - 180 = 3650$

Combine like terms

Group and simplify the constant numerical terms on the left-hand side: $2450 - 320 - 180 + 2y = 3650$ $2450 - 500 + 2y = 3650$ $1950 + 2y = 3650$

Isolate $y$

Subtract $1950$ from both sides: $2y = 3650 - 1950$ $2y = 1700$

Divide both sides by $2$ : $y = \frac{1700}{2} = 850\text{ mg}$

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student forgets to account for both aliquots and only adds y once instead of 2y, leading to an incorrect aliquot mass of 1700 mg.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
2.45 $\text{ g}$ = 2450 $\text{ mg}$ $\implies$ 0.18 $\text{ g}$ = 180 $\text{ mg}$ $\implies$ 3.65 $\text{ g}$ = 3650 $\text{ mg}$
Step 2
2450 - 320 + y - 180 = 3650
Correct
2450 - 320 + 2y - 180 = 3650
The problem states that two successive aliquots each containing $y\text{ mg}$ were added, so the total addition is $2y$ , not $y$ .
Step 3
1950 + y = 3650
Correct
1950 + 2y = 3650
You accurately simplified $2450 - 320 - 180 = 1950$ , but this step inherits the missing factor of $2$ on $y$ .
Step 4
y = 1700 $\text{ mg}$
Correct
2y = 1700 $\implies$ y = 850 $\text{ mg}$
Your subtraction $3650 - 1950 = 1700$ was calculated correctly, but dividing by the true coefficient $2$ yields $y = 850\text{ mg}$ .

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly converted all gram measurements to milligrams before combining them.

You correctly combined the constant numerical terms on the left side of the equation.

You applied valid algebraic isolation steps to find your final value.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Omitted Multiplier / Quantity Count

You represented the added solute as a single aliquot $+ y$ rather than accounting for both aliquots as $+ 2y$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Overlooking quantity keywords like 'two successive aliquots each containing $y$ ' when translating verbal descriptions into algebraic expressions.

REASONING PATH

Your ACEpath teacher’s suggested approach

Keyword Scan: Phrases like 'two aliquots each of mass $y$ ' mean total mass added $= 2 \times y$ .
Circle quantity adjectives ('two', 'each', 'twice') before writing the variable terms.
Dimension / Rate Check: $\text{Total added}$ = ( $\text{number of additions}$ ) $\times$ ( $\text{amount per addition}$ ).
Confirm that the coefficient in front of the variable matches the count of repeated additions.

CORRECTED PATH

ACEpath Teacher's Demonstration

Write the total mass added as $2y$ : $2450 - 320 + 2y - 180 = 3650$
Combine constant terms: $1950 + 2y = 3650$
Subtract $1950$ from both sides: $2y = 1700$
Divide by $2$ : $y = 850\text{ mg}$

Final answer

y = 850 $\text{ mg}$

NEXT STEPS

Final Takeaways from Your Teacher

Read word problems with a highlighter to catch multiplier words such as 'two successive', 'each', or 'per'.
Always define the variable term explicitly (e.g., 'let $2y$ = total mass of 2 aliquots') before inserting it into the balance equation.
Complete the three targeted practice exercises below to practice matching repeated additions to their correct algebraic coefficients.

OverviewACEpath Teacher's Diagnosis

02 The student misinterprets the final removal of 0.18 g as an addition, leading to an incorrect balance equation and finding y = 760 mg.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
2450 - 320 + 2y + 180 = 3650
Correct
2450 - 320 + 2y - 180 = 3650
You correctly converted the masses to milligrams ( $2.45\text{ g} = 2450\text{ mg}$ , $0.18\text{ g} = 180\text{ mg}$ , and $3.65\text{ g} = 3650\text{ mg}$ ) and represented the two aliquots as $+2y$ . However, the problem specifies that the chemist 'removes $0.18\text{ g}$ ', which requires subtracting $180\text{ mg}$ rather than adding it.
Step 2
2310 + 2y = 3650 $\implies$ 2y = 1340
Correct
1950 + 2y = 3650 $\implies$ 2y = 1700
Your algebraic simplification and subtraction across the equals sign were executed correctly based on your equation, but the numerical value was shifted by the earlier sign error.
Step 3
y = 670 $\text{ mg}$
Correct
y = 850 $\text{ mg}$
You correctly divided both sides by $2$ , but the final numerical value inherits the setup error from Step 1.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly converted all quantities from grams to milligrams using the factor $1\text{ g} = 1000\text{ mg}$ .

You accurately translated the two added aliquots into the algebraic term $+2y$ .

You followed sound algebraic steps to combine constants and isolate the variable.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Sign Error in Mass Balance Setup

You wrote $+ 180$ instead of $- 180$ when translating the removal of solute into the equation.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Conflating removals and additions during multi-step word problem transcription, causing a negative mass change to be assigned a plus sign in the balance equation.

REASONING PATH

Your ACEpath teacher’s suggested approach

Action-to-Sign Verification: Quantities being withdrawn, consumed, or removed must have a negative sign ( $-$ ), while additions and yields must have a positive sign ( $+$ ).
Read through each action verb in sequence and write down its explicit operational sign ( $+$ or $-$ ) directly beside the value before forming the full equation.

CORRECTED PATH

ACEpath Teacher's Demonstration

Write the corrected balance equation: $2450 - 320 + 2y - 180 = 3650$
Combine the constant terms on the left side: $(2450 - 320 - 180) + 2y = 3650 \implies 1950 + 2y = 3650$
Subtract $1950$ from both sides: $2y = 1700$
Divide by $2$ : $y = 850\text{ mg}$

Final answer

y = 850 $\text{ mg}$

NEXT STEPS

Final Takeaways from Your Teacher

Annotate word problems by writing a small $+$ above words like 'add' or 'deposit' and a small $-$ above words like 'withdraw', 'remove', or 'leak' before setting up your equation.
Double-check each term in your setup equation against the narrative flow to ensure every operation matches the physical process.

OverviewACEpath Teacher's Diagnosis

03 The student fails to convert the gram values to milligrams, directly mixing numerical values with mismatched units and obtaining y = 159.69 mg.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
2.45 - 320 + 2y - 0.18 = 3.65
Correct
2450 - 320 + 2y - 180 = 3650
You directly combined numerical values in grams ( $2.45\text{ g}$ , $0.18\text{ g}$ , $3.65\text{ g}$ ) with values in milligrams ( $320\text{ mg}$ , $2y\text{ mg}$ ). Because $1\text{ g} = 1000\text{ mg}$ , all masses must be converted to milligrams before writing the equation.
Step 2
-317.73 + 2y = 3.65 $\implies$ 2y = 321.38
Correct
1950 + 2y = 3650 $\implies$ 2y = 1700
Your algebraic manipulation was executed correctly according to your initial line, combining $2.45 - 320 - 0.18 = -317.73$ and adding $317.73$ to both sides, but the values are off due to the earlier unit mismatch.
Step 3
y = 160.69 $\text{ mg}$
Correct
y = 850 $\text{ mg}$
Your division step $\frac{321.38}{2} = 160.69$ was computed accurately, but the resulting mass is incorrect due to the initial unit error.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly identified the additive and subtractive actions in the balance equation.

You accurately represented the two identical aliquots as $2y$ .

You followed standard linear equation solving steps and computed the arithmetic accurately.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Unit Conversion Omission

You treated numerical values with different metric units (grams and milligrams) as if they were directly compatible.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Overlooking dimensional consistency before constructing an algebraic model, which leads to adding and subtracting numbers with a scale difference of $1000\times$ .

REASONING PATH

Your ACEpath teacher’s suggested approach

Unit Audit: Every term added or subtracted in a physical equation must share identical units.
Scan all given quantities and target variables before writing an equation, converting any grams to milligrams using $1\text{ g} = 1000\text{ mg}$ .

CORRECTED PATH

ACEpath Teacher's Demonstration

Convert each gram quantity: $2.45\text{ g} = 2450\text{ mg}$ , $0.18\text{ g} = 180\text{ mg}$ , and $3.65\text{ g} = 3650\text{ mg}$ .
Write the mass-balance equation: $2450 - 320 + 2y - 180 = 3650$ .
Combine constant terms on the left: $1950 + 2y = 3650$ .
Subtract $1950$ from both sides: $2y = 1700$ .

Final answer

y = 850 $\text{ mg}$

NEXT STEPS

Final Takeaways from Your Teacher

Write the unit next to each number during scratch work or highlight quantities that need metric prefix conversions.
Verify that both sides of an algebraic equation have consistent physical dimensions before simplifying.

Lock it in

Question

In a word problem, how do you translate 'added 3 successive portions each containing $x\text{ mL}$ ' into an algebraic term?

Answer

$+ 3x$ . Multiply the count of portions by the quantity per portion: $3 \times x = 3x$ .

Resources

Answer and steps

Final answer

Steps

Typical errors

01 The student forgets to account for both aliquots and only adds y once instead of 2y, leading to an incorrect aliquot mass of 1700 mg.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

02 The student misinterprets the final removal of 0.18 g as an addition, leading to an incorrect balance equation and finding y = 760 mg.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

03 The student fails to convert the gram values to milligrams, directly mixing numerical values with mismatched units and obtaining y = 159.69 mg.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

Lock it in

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