TPN Practice Problems

docx

School

University of Alabama *

*We aren’t endorsed by this school

Course

465

Subject

Chemistry

Date

Feb 20, 2024

Type

docx

Pages

Uploaded by KidKangaroo4464

TPN Practice Problems Your patient is receiving 1L of D 50 W per day. How many kcal and grams of dextrose are provided daily?  D 50 W = 50% dextrose  0.5 x 1000 ml = 500 g dextrose  Dextrose provides 3.4 kcal/g  500 g x 3.4 = 1700 kcal  So, if you gave a patient 1L of D 50 W, you would be giving 500 g dextrose and 1700 kcal. Your patient is receiving 1L of D 50 W per day. What is the osmolarity of the solution?  D 50 W = 50% dextrose  0.5 x 1000 ml = 500 g dextrose/L  Osmolarity = 5 x g/L dextrose  5 x 500 g/L = 2500 mOsm/L  So, D 50 W provides 500 g dextrose/L and 2500 mOsm/L Your patient is receiving 700 ml of D 50 W per day. How many kcal and grams of dextrose are provided daily?  D 50 W = 50% dextrose  0.5 x 1000 ml = 500 g dextrose in 1L  We are only giving 700 ml (or 0.7 L)  0.7 x 500 g = 350 g dextrose  Dextrose provides 3.4 kcal/g  350 g x 3.4 = 1190 kcal  So, if you gave a patient 700 ml of D 50 W, you would be giving 350 g dextrose and 1190 kcal. Your patient is receiving 700 ml of D 50 W per day. What is the osmolarity of this solution?

 D 50 W = 50% dextrose  0.5 x 1000 ml = 500 g dextrose in 1L  Osmolarity = 5 x g/L dextrose  5 x 500 g/L = 2500 mOsm/L  Note: Osmolarity is calculated as mOsm/L, so you aren’t concerned with how many liters you’re giving; you’re concerned with g/L. Your patient is receiving 500 ml of D 20 W per day. How many kcal and grams of dextrose are provided daily?  D 20 W = 20% dextrose  0.2 x 1000 ml = 200 g dextrose in 1L  We are only giving 500 ml (or 0.5 L)  0.5 x 200 g = 100 g dextrose  Dextrose provides 3.4 kcal/g  100 g x 3.4 = 340 kcal  So, if you gave a patient 500 ml of D 20 W, you would be giving 100 g dextrose and 340 kcal. Your patient is receiving 500 ml of D 20 W per day. What is the osmolarity of this solution?  D 20 W = 20% dextrose  0.2 x 1000 ml = 200 g dextrose in 1L  Osmolarity = 5 x g/L dextrose  5 x 200 g/L = 1000 mOsm/L  Note: Osmolarity is calculated as mOsm/L, so you aren’t concerned with how many liters you’re giving; you’re concerned with grams/L. Your patient is receiving D 70 W at 50 ml/hr. How many kcal and grams of dextrose are they receiving per day?  D 70 W = 70% dextrose  0.7 x 1000 ml = 700 g dextrose in 1L  We are giving 50 ml/hr x 24 hrs = 1200 ml (or 1.2 L)

 1.2 L x 700 g/L = 840 g dextrose  Dextrose provides 3.4 kcal/g  840 g x 3.4 = 2856 kcal  So, if you gave a patient 1.2 L of D 70 W, you would be giving 840 g dextrose and 2856 kcal. Your patient is receiving D 70 W at 50 ml/hr. What is the osmolarity of the solution?  D 70 W = 70% dextrose  0.7 x 1000 ml = 700 g dextrose in 1L  Osmolarity = 5 x g/L dextrose  5 x 700 g/L = 3500 mOsm/L  Note: Osmolarity is calculated as mOsm/L, so you aren’t concerned with how many liters you’re giving, you’re concerned with g/L. Your patient is receiving 500 ml of 10% AA per day. How many kcal and grams of AA are they receiving daily?  10% AA  0.1 x 1000 ml = 100 g AA in 1 L  We are giving 500 ml/day, so 0.5 L  0.5 L x 100 g/L = 50 g AA  AA provides 4 kcal/g  50 g x 4 = 200 kcal from AA  So, if you gave a patient 0.5 L of 10% AA, you would be giving 50 g AA and 200 kcal . Your patient is receiving 500 ml of 10% AA per day. What is the osmolarity of the solution?  10% AA  0.1 x 1000 ml = 100 g AA in 1 L  Osmolarity = 10 x g/L AA  10 x 100 g/L = 1000 mOsm/L

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

Your patient is receiving 8.5% AA at 50 ml/hr. How many kcal and grams of protein are they receiving per day?  8.5% AA  0.085 x 1000 ml = 85 g AA in 1L  We are giving 50 ml/hr x 24 hrs = 1200 ml (or 1.2 L)  1.2 L x 85 g/L = 102 g AA  AA provides 4 kcal/g  102 g x 4 = 408 kcal from AA  So, if you gave a patient 1.2 L of 8.5% AA, you would be giving 102 g AA and 408 kcal. Your patient is receiving 8.5% AA at 50 ml/hr. What is the osmolarity of the solution?  8.5% AA  0.085 x 1000 ml = 85 g AA/L  Osmolarity = 10 x g/L AA  10 x 85 g/L = 850 mOsm/L Your patient is receiving 250 ml of 20% lipid per day. How many kcal and grams of lipid does this provide?  20% lipid  0.2 x 1000 ml = 200 g/L  We are giving 250 ml, so 0.25 L  200 g/L x 0.25 L = 50 g lipid  20% lipid provides 10 kcal/g  50 g x 10 kcal/g = 500 kcal Your patient is receiving 250 ml of 20% lipid per day. What is the osmolarity of this solution?

 N/A – lipid does NOT contribute to osmolarity Your patient is receiving 500 ml of 10% lipid today. How many kcal and grams of lipid does this provide?  10% lipid  0.1 x 1000 ml = 100 g/L  We are giving 500 ml, so 0.5 L  100 g/L x 0.5 L = 500 g lipid  10% lipid provides 11 kcal/g  50 g x 11 kcal/g = 550 kcal Your patient is receiving D 50 W and 10% AA at 85 ml/hr. How many kcal, grams of protein, and grams of CHO are provided daily?  D 50 W = 50% dex = 0.5 x 1000 ml = 500 g/L  10% AA = 0.1 x 1000 ml = 100 g/L  85 ml/hr x 24 hrs = 2040 ml, or 2.04 L  2.04 L x 500 g/L dex = 1020 g dextrose  2.04 L x 100 g/L AA = 204 g AA  1020 g dextrose x 3.4 kcal/g = 3468 kcal/dex  204 g AA x 4 kcal/g = 816 kcal/AA  Total kcal: 3468 + 816 = 4284 kcal Your patient is receiving D 50 W and 10% AA at 85 ml/hr. What is the osmolarity of the solution?  50% dex = 500 g/L  500 g/L x 5 = 2500 mOsm/L  10% AA = 100 g/L  100 g/L x 10 = 1000 mOsm/L

 Total osmolarity: 2500 + 1000 = 3500 mOsm/L Your patient is receiving D 50 W and 10% AA at 85 ml/hr. Could this solution be given via PPN?  Total osmolarity: 2500 + 1000 = 3500 mOsm/L  No , the solution is > 900 mOsm/L, so it could NOT be given via PPN. It could only be given via central TPN. Your patient is receiving 450 ml of D 50 W and 600 ml of 8.5% AA per day. How many grams of dextrose and AA is he receiving daily?  D 50 W = 50% dextrose = 500 g/L  500 g/L x 0.45 L = 225 g dex  8.5% AA = 0.85 x 1000 ml = 85 g/L  85 g/L x 0.6 L = 51 g AA Your patient is receiving 450 ml of D 50 W and 600 ml of 8.5% AA per day. How many kcal is he receiving daily?  225 g dex x 3.4 kcal/g = 765 kcal dextrose  51 g AA x 4 kcal/g = 204 kcal AA  Total kcal: 765 + 204 = 969 kcal Your patient is receiving 450 ml of D 50 W and 600 ml of 8.5% AA per day. What is the osmolarity of the solution? Should it be given as TPN or PPN?  D 50 W = 50% dex = 500 g/L  500 g/L x 5 = 2500 mOsm/L  8.5% AA = 0.085 x 1000 ml = 85 g/L  85 g/L x 10 = 850 mOsm/L  Total Osmolarity: 2500 + 850 = 3350 mOsm/L

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

 This CANNOT be given via PPN. It should be given via TPN. o For PPN, it should be ≤ 900 mOsm/L Your patient weighs 75 kg. How many grams of dextrose can she tolerate per day (using 3 – 5 mg/kg/min)?  3 mg/kg/min – 5 mg/kg/min  There are 1440 min/day (24 hrs x 60 min)  3 x 75 x 1440 = 324000 mg CHO = 324 g  5 x 75 x 1440 = 540000 mg CHO = 540 g  So, she can tolerate 324 – 540 g dex/day Your patient needs 125 g protein per day. Your pharmacy stocks 12% AA. How many ml of 12% AA does she need per day to meet her needs?  12% AA = 120 g/L  125 g ÷ 120 g/L = 1.042 L  1.042 L = 1042 ml  Your patient needs 1042 ml of 12% AA to provide her with 125 g protein/day. Your patient needs 90 g protein per day. Your pharmacy stocks 10% AA. How many ml of 10% AA does she need per day to meet her needs?  10% AA = 100 g/L  90 g ÷ 100 g/L = 0.9 L  0.9 L = 900 ml  Your patient needs 900 ml of 10% AA to provide her with 90 g protein/day. A TPN bag with 1500 ml has a final concentration of 8% dextrose and 3.2% AA. How many grams of dextrose and AA does the bag contain?  8% dextrose = 0.08 x 1000 ml = 80 g/L dex  3.2% AA = 0.032 x 1000 ml = 32 g/L AA  We are giving 1500 ml, or 1.5 L  80 g/L x 1.5 L = 120 g dextrose  32 g x 1.5 L = 48 g AA

A TPN bag with 1500 ml has a final concentration of 8% dextrose and 3.2% AA. How many kcal does the bag provide?  80 g/L x 1.5 L = 120 g dextrose  120 g x 3.4 kcal/g = 408 kcal from dex  32 g/L x 1.5 L = 48 g AA  48 g x 4 kcal/g = 192 kcal from AA  Total kcal: 408 + 192 = 600 kcal. A TPN bag with 1500 ml has a final concentration of 8% dextrose and 3.2% AA. What is the osmolarity of the solution? Can you give via PPN?  80 g/L dex x 5 = 400 mOsm/L  32 g/L AA x 10 = 320 mOsm/L  Total Osmolarity: 400 + 320= 820 mOsm/L  Yes , you can give this via PPN, as the total osmolarity is ≤ 900 mOsm/L. Your patient who weighs 60 kg needs 1800 kcal & 60 g protein/day. Your pharmacy stock D 50 W, 10% AA, and 20% lipid. Write a TPN prescription to meet his needs.  Step 1: Determine how many ml of AA to provide. o 10% AA provides 100 g/L o Your patient needs 60 g o 60 g ÷ 100 g/L = 0.6 or 600 ml o Your patient needs 600 ml of 10% AA  Step 2: Determine how many grams dex to provide. (using 3 – 5 mg/kg/min) o 3 mg/kg/min  3 x 60 x 1440 = 259200 mg of CHO, or 259.2 g CHO o 5 mg/kg/min  5 x 60 x 1440 = 432000 mg of CHO, or 432 g CHO o So, your patient needs ~259 – 432 g CHO/day o Let’s go in the middle and aim for 350 g dextrose/day  It’s also correct as 259 + 432 = 691 / 2 = 345.5 = 346 g dex/day

 Step 3: Determine how many ml of dex to provide. o We want to give 350 g dex/day o D 50 W provides 500 g/L o 350 g ÷ 500 g/L = 0.7 L or 700 ml D 50 W  Step 4a: Calculate the remaining kcal to provide from lipid. o We’re giving 60 g AA and 350 g CHO/day o 60 g AA x 4 kcal/g = 240 kcal AA o 350 g CHO x 3.4 kcal/g = 1190 kcal dextrose o 1800 kcal total – 240 – 1190 = 370 kcal left to give from lipid  Step 4b: Make sure this is < 30% of kcal. o 370 kcal from lipid ÷ 1800 kcal total = 0.2056 = 20.56% kcal from lipid o We are under 30%, so we are good  Step 5: Determine how many ml of lipid to provide. o 20% lipid provides 10 kcal/g o 370 kcal of lipid ÷ 10 kcal/g = 37 g lipid o 20% lipid provides 200 g/L o We need to provide 37 g o 37 g ÷ 200 g/L = 0.185 L or 185 ml of 20% lipid  Our final TPN prescription is: o 600 ml of 10% AA o 700 ml of D 50 W o 185 ml of 20% lipid o Let’s verify how much this provides  10% AA = 100 g/L x 0.6 L = 60 g AA ✅  60 g x 4 kcal/g = 240 kcal from AA  50% dex = 500 g/L x 0.7 L = 350 g dex ✅  350 g x 3.4 kcal/g = 1190 kcal dex  20% lipid = 200 g/L x 0.185 L = 37 g lipid ✅  37 g x 10 kcal/g = 370 kcal lipid  Total kcal: 240 + 1190 + 370 = 1800 kcal ✅ Our TPN prescription is: 600 ml of 10% AA, 700 ml of D 50 W, 185 ml of 20% lipid. What is the final concentration of this solution?

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

 Total volume: 600 ml + 700 ml + 185 ml = 1485 ml  60 g AA / 1485 ml = 0.0404 = 4.04% AA  350 g dex / 1485 ml = 0.2357 = 23.57% dex  37 g lipid / 1485 ml = 0.0249 = 2.49% lipid